KSEEB Solutions For Class 9 Science Chapter 3 Atoms And Molecules

KSEEB Solutions For Class 9 Science Chapter 3 Atoms and Molecules Important Concepts

Atoms and molecules, law of constant proportions, atomic and molecular masses, mole concept, valency and chemical formula of common compounds.

Atom
It is the smallest particle which takes part in chemical reaction. Example: Na, K, Fe, etc.

Molecule
It is made up of two or more atoms and it is capable to exist independently. Example: H2,Cl2,O2 etc

Atomicity
The number of atoms present in one molecule of a substance is called its atomicity.
Examples: Monoatomic, diatomic, triatomic, tetra atomic.

Read and Learn More KSEEB Solutions for Class 9 Science 

Ion
An atom or group of atoms which carries positive or negative charge is called an ion.

Ionic compounds
The compounds consisting of cations and anions are called ionic compounds.

Cation
It is a positively charged ion.
Example: Na+, K+, Ca2+, Mg2+ etc

Anion
It is a negatively charged ion.
Example: Cl , Br~, F~, O2” etc

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Monoatomic
They consist of only one atom.
Example: H, He, S, etc

Diatomic
They consist of two atoms. Example: H2,02, N2, etc.

Triatomic
They consist of three atoms. Example: O3

Tetra atomic
They consist of four atoms. Example: P4

Law of conservation of mass
It states that matter can neither be created nor be destroyed in a chemical reaction.

Dalton’s atomic theory
All matter whether element, compound or mixture is composed of small particles called atoms.

Atomic mass unit
It is defined as the mass of 1/12th of the mass of 1 atom of car bon-12 isotope.

Atoms And Molecules KSEEB Class 9 Question Answers 

Mole concept
Mole is counting unit for atoms, molecules or ions and is equal to 6.022 x1023.
This number is called Avogadro’s constant and is denoted by symbol N.

Mole
The amount of a substance which contains as many particles (atoms, ions or molecules) as in 12 g of C-12.  1 mole = 6.022 x -1023  atoms.

Gram atomic mass
Atomic mass of a substance expressed in grams is called its gram atomic mass. This amount is called one gram atom.

Gram molecular mass
Molecular mass of a substance expressed in grams is called gram molecular mass. This amount is called one gram molecule.

Valency of an element
The combining capacity of the element.

Atoms And Molecules Exercises

Question 1. A 0.24g sample of compound of Oxygen and Boron was found by analysis to contain 0.096g of Boron and 0.144g of Oxygen. Calculate the percentage
composition of the compound by weight.
Answer Percentage of any element in a compound Mass of the element
\(=\frac{\text { Mass of the element }}{\text { Mass of the compound }} \times 100\)

Percentage of Oxygen
\(=\frac{\text { Mass of Oxygen }}{\text { Mass of compound }} \times 100\)
\(=\frac{0.144}{0.24} \times 100=60 \%\)

Percentage of boron
\(=\frac{\text { Mass of Boron }}{\text { Mass of compound }} \times 100\)

Question 2. When 3.0g of Carbon is burnt in 8.0g of Oxygen, ll.OOg of Carbon dioxide is produced. What mass of Carbon dioxide will be formed when 3.00g of Carbon is burnt in 50.00g of Oxygen? Which law of chemical
combination will govern you answer?
Answer 3g of Carbon produce Carbon dioxide=l 1 g The remaining oxygen 50g – 8g = 42g does not take part in the reaction. The law of definite proportion is governed by the above data.

Question 3. What are polyatomic ions? Give examples.
Answer Polyatomic ions: Two or more different atoms unite to form a charged particle and are called polyatomic ions.
Example: (PO4)-3 Phosphate, Nitrate(NG3)1

Question 4. Write the chemical formulae of the following,
1. Magnesium chloride
Answer
\(\mathrm{Mg}^{2+}+\mathrm{Cl}^{-1} \rightarrow \mathrm{Mg}^{2+} \mathrm{Cl}^{-1}\)

2. Calcium oxide
Answer
\(\mathrm{Ca}^2+\mathrm{O}^2 \rightarrow \mathrm{Ca}^{2+} \mathrm{O}^{2-}\)

3. Copper nitrate
Answer

\(\mathrm{Cu}^{2+}+\mathrm{NO}_3^{-1} \rightarrow \mathrm{Cu}^{2+}+\mathrm{NO}_3^{-1}\)

4. Aluminum chloride
Answer
\(\mathrm{Al}^{3+}+3 \mathrm{Cl} \rightarrow \mathrm{Al}^{3+} \mathrm{Cl}^{-1}\)

5. Calcium carbonate
Answer
\(\mathrm{Ca}^{2+}+\mathrm{CO}_3^{2-} \rightarrow \mathrm{Ca}^{2+} \mathrm{CO}_3^{2-}\)

Question 5. Give the names of the elements present in the following compounds:
1. Quicklime
Answer CaO – Elements present are calcium and oxygen

2. Hydrogen Bromide
Answer HBr – Elements present are hydrogen and Bromide

3. Baking powder
Answer NaHC03 – Elements present are Sodium hydrogen carbon and oxygen

4. Potassium Sulphate
Answer
\(\mathrm{K}_2 \mathrm{SO}_4-\)Elements present are potassium, sulphur and oxygen

KSEEB Solutions For Class 9 Science Chapter 3 Atoms And Molecules

Laws Of Chemical Combination Class 9 KSEEB Solutions 

Question 6. Calculate the molar mass of the following substances.
1. Ethyne \(\mathrm{C}_2 \mathrm{H}_2\)
2. Sulphur molecule S8
3. Phosphorus molecule P4(Atomic Mass of P=31) Hydrochloric acid, HCl
Nitric acid \(\mathrm{HNO}_3\)
Answer:
1. Molecular mass of \(\mathrm{C}_2 \mathrm{H}_2\) (Ethyne)
= 2 x atomic mass of C + 2 x atomic mass of H
= 2 x 12 + 2 x 1 =26g

2. Molecular mass of S (Sulphur)
= 8 x atomic mass of S
= 8 x 32 = 256g

3. Molecular mass of P4 (Phosphorus)
= 4 x atoinic mass of P
= 1×31 = 124g

4. Molecular mass of HCl (Hydrochloric acid) = 1 x atomic mass of H + 1 x atomic mass of Cl
= 1 x 1 +35,5 = 36.5g

5. Molecular mass of HN03 (Nitric acid)
= 1 x atomic mass of H + 1 x atomic mass of
N + 3 x atomic mass of O
= 1 X 1 + 1 X 14 + 3 x 16 = 63g

Question 7. What is the mass of
1. 1 mole of Nitrogen atom
2. 4 moles of Aluminium atom (Atomic mass of A1 = 27)
3. 10 moles of Sodium Sulphite \(\left(\mathrm{Na}_2 \mathbf{S O}_3\right)\)

Answer:
1. 1 mole of nitrogen atom
= 1 x gram atomic mass of nitrogen atom
= 1 x 14= 14g

2. 4 moles of Aluminium atoms
= 4 x gram atomic mass of Aluminium atoms
= 4x 27= 108g

3. 10 moles of Sodium Sulphite \(\left(\mathrm{Na}_2 \mathbf{S O}_3\right)\)
= 10 (2 x gram atomic mass of Na + 1 x gram 1
atomic mass of Sulphur + 3 x gram atomic mass of oxygen)
= 10 (2 x 23 + 1 x 32 + 3 x I6)g
= 10 (46g + 32g + 48g)
= 10x 126g=1260g

Question 8. Convert into mole
1. 12g of Oxygen gas
2. 20g of Water
3. 22g of Carbon dioxide
Answer
1. Number of moles \((\mathrm{n})=\frac{\text { Given Mass }(\mathrm{m})}{\text { Molar Mass }(\mathrm{M})}\)
Molecular mass of O2 = 2’x 16 = 32g
32g o f Oxygen = 1 mole
\(12 \mathrm{~g} \text { of Oxygen }=\frac{12}{32} \text { mole }\)
= 0.375 mole

2. Molar mass of \(\mathrm{H}_2 \mathrm{O}\) = 2 x Igx 1 x 16g= 18g
18g of Water = 1 mole
\(20 \mathrm{~g} \text { of Water }=\frac{20}{18} \text { mole }\)
= 1,11 mole

3. Molar mass of \(\mathrm{CO}_2\)= 1 x 12g + 2 x 16g = 44g
44g of Carbon dioxide = 1 mole
\(22 \mathrm{~g} \text { of Carbon dioxide }=\frac{22}{44} \text { mole }\)
= 0.5 mole

Question 9. What is the mass of
1.0.2 mole of Oxygen atoms
2. 0.5 mole of Water Molecules
Answer
1. 1 mole of Oxygen atoms
= 1 x 16= 16g
0.2 mole of Oxygen atoms
= 16g x .0.2 = 3.2g

2. 1 mole of Water (H20) molecules
=2 x lg+ 1 x igg = 18g
0.5 mole of Water \(\left(\mathrm{H}_2 \mathrm{O}\right)\)molecules
= 18g x 0.5 = 9.0g

Question 10. Calculate the number of molecules of Sulphur
\(\left(S_8\right)\)present in 16g of solid sulphur.

Answer
\(\text { Number of moles }=\frac{\text { Given Mass }(\mathrm{m})}{\text { Molar Mass }(\mathrm{M})}\)
1 mole of Sulphur (Sx)
= 8><1 gram atomic mass of Sulphur
= 8 x 32 g = 256g
1 mole of S„ molecules
= 6.022 x 1023 molecules
256g of molecules has
= 6.022 x \(10^{23}\)molecules
16g of molecules has
\(=\frac{6.022 \times 10^{23} \times 16}{256} \text { molecules }\)
= 3.76x 1022moiecules.

Dalton’s Atomic Theory KSEEB Class 9 Notes 

Question 11. Calculate the number of Aluminium ions
present in 0.051g of Aluminium Oxide \(\left(\mathrm{Al}_2 \mathrm{O}_3\right)\) (Hint: The mass of an iron is the same as that of an atom of the same element. Atomic mass of AI = 27 u.)
Answer
1 gram molecule (1 mole) of \(\mathrm{Al}_2 \mathrm{O}_3\)
= 2 x gram atomic mass of Al + 3 x gram
atomic mass of O
= 2 x 27g + 3 x I6g
= 54g+48g= 102g
1 gram molecule (1 mole) of A1203 contains Aluminum ions
= 2 x 6.022 x 1023
= 12.044 x 1023
102g of Al2 O 3 has number of Aluminium ions
= 12.044 x 1023
0.05 lg of has number ofAluminium ions
\(=\frac{12.044 \times 10^{23} \times 0.051}{102}\)

Atoms And Molecules Textual Questions

Question 1. In a reaction, 5.3g of Sodium Carbonate reacted with 6g of Ethanoic acid. The products where 2.2g of Carbon dioxide, 0.9g of water and 8.2g of Sodium Ethanoate. Show that these observations are in agreement with the law of conservation of mass. Sodium Carbonate + Ethanoic acid sodium > Ethanoate + Carbon dioxide + Water
Answer
Total mass of the reaction equals of Sodium Carbonate + Mass of Ethanoic acid solution.
= 5.3g + 6g= 11.3g
Total mass of products = Mass of sodium ethanoate solution + Mass of carbon dioxide
= 8.2g + 2.2g+0.9g=U.3g
The mass of reactants is equal to the mass of products, therefore, it proves law of conservation of mass.

Question 2. Hydrogen and oxygen combine in the ratio 1: 8 by mass to form water. What mass of Oxygen gas would be required to react completely with 3g of hydrogen gas?
Answer As hydrogen and oxygen combine in the ratio 1: 8 by mass, this means that lg of hydrogen combines with 8g of oxygen.
3g of hydrogen will react with oxygen
= 8 x 3g = 24g

Question 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer The following postulate of Dalton’s atomic theory is a result of the law of conservation of mass. “Atoms are indivisible particles which cannot be created or destroyed in a chemical reaction.”

Question 4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer “The relative number and kinds of atoms are constant in a given compound.”

Question 5. Define the atomic mass unit.
Answer Themassof l/12ftpartof C-12 is equivalent to one atomic mass unit.

Question 6. Why is it not possible to see an atom with. naked eyes?
Answer Because an atom is too small the atomic radii of an atom is of the order 10’10mtol0’9m

Question 7. Write down the formulae of the following.
1. Sodium oxide
2. Aluminium chloride
3. Sodium sulphide
4. Magnesium hydroxide

Answer
1. Sodium oxide —>\(\mathrm{Na}_2 \mathrm{O}\)
2. Aluminum chloride —>\( \mathrm{AlCl}_3\)
3. Sodium sulphide—>\(\mathrm{Na}_2 \mathrm{~S}\)
4. Magnesium hydroxide —>\(\mathrm{Mg}(\mathrm{OH})_2\)

Question 8. Write the names of the compounds represented by the following formulae.
1\(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\)
2.\(\mathrm{CaCl}_2\)
3.\(\mathrm{K}_2 \mathrm{SO}_4\)
4.\(\mathrm{KNO}_3\)
5.\(\mathrm{CaCO}_3\)

Answer

  1. \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\)—>Aluminium Sulphate
  2. \(\mathrm{CaCl}_2\)—> Calcium Chloride
  3. \(\mathrm{K}_2 \mathrm{SO}_4\)—>Potassium Sulphate
  4. \(\mathrm{KNO}_3\)—> Potassium Nitrate
  5. \(\mathrm{CaCO}_3\) —> Calcium Carbonate

Question 9. What is meant by the term chemical formula?
Answer The chemical formula of a compound is a symbolic representation of its composition and actual number of atoms in one molecule of a pure substance, maybe an atom or a compound.

Question 10. How many atoms are present in a
1. \(\mathrm{H}_2 \mathrm{~S}\)molecule
2. \(\mathrm{PO}_4^{3-}\)ion

Answer
1. \(\mathrm{H}_2 \mathrm{~S}\)3 atoms are present.
2. \(\mathrm{PO}_4^{3-}\)5 atoms are present,

Mole Concept KSEEB Class 9 Textbook Solutions 

Question 11. Calculate the molecular mass of Molecular mass of\( \mathrm{H}_2, \mathrm{O}_2, \mathrm{Cl}_2\) \(\mathrm{CO}_2, \mathrm{CH}_4, \mathrm{C}_2 \mathrm{H}_6, \mathrm{C}_2 \mathrm{H}_4, \mathrm{NH}_3, \mathrm{CH}_3 \mathrm{OH}\)
Answer
Molecular mass of
\(\mathrm{H}_2=2 \times\)atomic mass of H
= 2 x lg = 2g
\(\mathrm{O}_2\)= 2 x 16g = 32g
\(\mathrm{Cl}_2\)= 2 x 35.5 = 71 g
\(\mathrm{CO}_2\)-1 x 12g + 2 x 16g = 44g
\(\mathrm{CH}_4\)– 1 x 12g + 4 x 1 g = 16g
\(\mathrm{C}_2 \mathrm{H}_6\) = 2X 12g + 6x lg = 30g
\(\mathrm{C}_2 \mathrm{H}_4\) = 2X 12g + 4x lg = 28g
\(\mathrm{NH}_3\)= 1 x 14g + 3 x lg= 17g
\(\mathrm{CH}_3 \mathrm{OH}\)= l X 12g + 4 X lg+ lx 16g = 32g

Question 12. Calculate the formula unit masses of ZnO, Na20, K2C03. Given atomic masses of (Zn 65u, Na = 23u, K = 39u, C = 12u and 0 = 16 u)

Answer The formula unit mass of ZnO
= 1 x atomic mass of Zn x 1 x atomic mass of C
= 1 x 65 + 1 x 16 = 81u
The formula unit mass of\(\mathrm{Na}_2 \mathrm{O}\)
= 2x 23 + 1x 16 = 62u
The formula unit mass of\(\mathrm{K}_2 \mathrm{CO}_3\)
= 2x 39+1 x 12 + 3x 16 = 138u

Question 13. If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?

Answer
l mole (6.022 x10^{23}  atoms) of Carbon weighs
= 12g
1 atom of 1 carbon weighs =\(\frac{12}{6.022 \times 10^{23}}\)
=1.99 x 10’23g

Question 14. Which has more number of atoms, 100 grams of Sodium or 100 grams of Iron?
(Given atomic mass of Na = 23u, Fe = 56u)

Answer 23g of sodium has atoms = N
100g of Sodium has atoms =\( \frac{100}{23} \mathrm{~N}_0\)
= 4.3 \(\mathrm{N}_0\)
56g of Iron has atoms = N
100 g of Iron has atoms =\(\frac{100}{56} \mathrm{~N}_0\)
= 1.78 N.
1oog of Sodium has more atoms than l00g of Iron.

Atoms And Molecules Additional Questions

Question 1. Classify each of the following on the basis of their atomicity.

1. \(\mathrm{NO}_2\)
2. \(\mathrm{He}\)
3. \(\mathrm{H}_2 \mathrm{O}_2\)
4. \(\mathrm{CH}_4\)

Answer:
1. Triatomic
2. Monoatomic
3. Tetra atomic
4. Renta atomic

Question 2. What do you understand by 1 amu or lu?
Answer 1 amu or lu stands for 1/12th of the mass of an atom of Carbon-12 isotope.

Question 3. What is the difference between an atom and a molecule?
Answer Atom is the smallest particle of an element that may or may not be capable of free existence. Molecule is the smallest particle of an element or compound which is capable of free existence.

Question 4. What do you understand by the ‘atomicity’ of a substance?
Answer: The number of atoms present in 1 molecule of a substance is called its atomicity.

Question 5. A glass jar contains 1.7g of Ammonia gas. Calculate the following:
1. Molar mass of Ammonia
2. How many moles of ammonia are present in the glass jar?

Answer:
1. Molar mass of Ammonia = 14 + 3 = 17g
2. No of moles \(=\frac{\text { Given Mass }(\mathrm{m})}{\text { Molar Mass }(\mathrm{M})}=\frac{1.7}{17}\)
\(=\frac{1}{10}=0.1\)mole

Atoms And Molecules High-order Thinking Questions

Question 1. Which of the following species is electrically neutral and why? \(\mathbf{K}^{+}, \mathbf{C l}^{-}, \mathbf{S}^{2-}, \mathbf{K}\)
Answer K, because it has equal number of protons and neutrons.

Question 2. Give two examples to show that law of conservation of mass applies to physical change also.

Answer Law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. This law applies to physical changes.
Example: Melting of wax, melting of ice into water.

Question 3. Calculate the molecular mass of the following compounds.
\({ a. } \mathrm{H}_2 \mathrm{SO}_4\)

\({ b. } \mathrm{Na}_2 \mathrm{CO}_3\)

Answer
1. \(\mathrm{H}_2 \mathrm{SO}_4\)=2 x H + 1 x s I 4 x O
= 2 x 1 + 1 x 32 + 4 x 16-98u
\(\mathrm{Na}_2 \mathrm{CO}_3=\)2xNa+lxC + 3xO
= 2X23+1X]2 + 3X 16 = 106u

Question 4. Sample of Vitamin C is known to contain 2.58 x 1024 oxygen atoms. How many moles of oxygen atoms are present in the sample?
Answer: 1 mole of Oxygen atom = 6.023 x \(10^{23}\) atoms
Number of moles of oxygen atoms
\(=\frac{2.58 \times 10^{24}}{6.022 \times 10^{23}}=4.28\) mole
=4.28 moles of oxygen atoms.

Chemical Formulas KSEEB Science Class 9 Chapter 3 

Atoms And Molecules Unit Test Multiple Choice Questions

Question 1. 3.42g of sucrose is dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution ______

1. 6.68 x\(10^{23}\)
2. 6.09x \(0^{22}\)
3. 6.022 x\(0^{23}\)
4. 6.022\(\times 10^{21}\)

Answer (1)

Question 2. The balancing of chemical equation is based on______

  1. Law of conservation of mass
  2. Law of combining value
  3. Law of constant proportion
  4. Avogadro’s law

Answer (1)

Question 3. Which of the following contains maximum number of molecules?______

  1. lg\(\mathrm{CO}_2\)
  2. lg\(\mathrm{N}_2\)
  3. lg\(\mathrm{H}_2[/atex]
  4. Ig[latex]\mathrm{CH}_4l[/atex]

Answer (3)

Atoms And Molecules True Or False

1. Sodium shows valency of [latex]\mathrm{Na}^2\) and\(\mathrm{Na}^{2+}\) False
2. Isotopes have same atomic mass. False
3. CO is carbon monoxide while CO is cobalt. True

Atoms And Molecules Answer The Following

Question 1.  What is lu equal to?
Answer of mass of 1 atom ofC-12.

Question 2. Name the element used in filament of electric bulb.
Answer Tungsten

Question 3. An element X shows variable valencies of 3 and 5. What will be the formulae of its oxides?

Answer
\({ }_3^{\mathrm{X}}=\mathrm{X}_2 \mathrm{O}_3\)
\(\bar{X}_5^{\mathrm{X}}=\mathrm{X}_2 \mathrm{O}_5\)

Question 4. Calculate the molecular mass of the following substances:
1. Ethyne \mathrm{C}_2 \(\mathrm{H}_2\)
2. Sulphur \(S_8\)

Answer
1. Ethyne \mathrm{C}_2 \(\mathrm{H}_2\)
Molecular mass of \(\mathrm{C}_2 \mathrm{H}_2\)= 2 x atomic mass of
C + 2 x atomic mass of H
= 2 xl2u + 2 x lu = 26u
2. Sulphur\(\mathrm{S}_8\)
Molecular mass of = 8 x atomic mass of Sulphur
= 8 x 32u =256u

Question 5. What is the difference between 2H and H2
Answer 2H represents two atoms of hydrogen and represents a molecule of hydrogen

Atoms And Molecules Answer The Following

Question 1. State the rules for writing the chemical formulae.

Answer

  1. The valency or changes on the iron must be balanced
  2. When a compound consists of a metal and a non-metal, the name or symbol of the metal is written first.
  3. In compounds formed with polyatomic ions, the iron is enclosed in a bracket before writing the number to indicate the ratio.

Question 2. Nagraj took 5 moles of carbon atoms in a container and Ramesh also took 5 moles of sodium atoms in another container of same weight
1. Who’s container is heavier?
2. Whose container has more number of atoms?

Answer
1. Mass of sodium atoms carried by Nagraj = (5 x23)g=115g
While mass of carbon atoms carried by Ramesh = (5 x 12)g = 60g
Thus, Nagraj container is heavy.
2. Both the containers have same number of atoms as they have same number of moles of atoms.

Numerical Problems On Mole Concept Class 9 KSEEB 

Question 3. Write the cations and anions present in the following compounds.

Answer
Compound
1. \(\mathrm{CH}_3 \mathrm{COONa}\)
2. NaCl
3. \(\mathrm{NH}_4 \mathrm{NO}_3\)

Cation
1.\(\mathrm{Na}^{+}\)
2. \(\mathrm{Na}^{+}\)
3.\(\mathrm{NH}_4^{+}\)

Anion
1.\(\mathrm{CH}_3 \mathrm{COO}^{-}\)
2.\(\mathrm{Cl}^{-}\)
3.\(\mathrm{NO}_3^{-}\)

Question 4. State the atomicity of the following molecules,

Answer:

  1. Oxygen —» Diatomic
  2. Phosphorus —» Tetra atomic
  3.  Sulphur  —» Polyatomic
  4. Argon —> Monatomic

Atoms and molecules Activity

Question 1. The Law of Conservation of Mass.

 

Law of Conservation

 

Atoms And Molecules Experimental Procedure


Prepare 5% solution of sodium sulphate and Barium Chloride. A little amount of sodium sulphate solution is taken in a conical flask and a little solution of Barium Chloride is taken in an ignition tube. Hang the ignition tube in the conical flask carefully as shown in the figure. Put a cork on the flask and weigh the flask with its contents. Now tilt and swirl the flask so that the two solutions mix up well. Weigh the flask again.

Observation: Chemical reaction takes place. Weight of the flask remains same before and after the reaction.
Conclusion: Total mass o f the products = Total mass of the reactants. Thus, the law of conservation of mass is proved.

KSEEB Solutions For Class 9 Science Chapter 9 Force And Laws Of Motion

KSEEB Solutions For Class 9 Science Chapter 9 Force And Laws Of Motion Important Concepts

Force: It is a push or pull acting on a body and its S.I un it is kgms’2 or newton
Newton: It is a force that produces an acceleration of 1 ms-2on a body of mass one kg.
Balanced forces: Two or more forces whose resultant is zero
Unbalanced forces: Two or more forces whose resultant is not zero
Newton’s first law of motion: Abody in the state of rest or motion will remain so unless acted upon by an unbalanced force.
Inertia: The natural tendency of a body to change its state of rest or uniform motion by itself Mass: It is a measure of inertia of a body and its S.I unit
Types of inertia: Inertia rest, the inertia of motion, and inertia of direction
Momentum: It is the amount of motion possessed by a moving body. It is measured as the product of the mass and velocity of the body. S.I unit is kgms-1
Newton’s second law: The rate of change of momentum of a body is directly proportional to the force and takes place in the direction of the force.
Newton’s third law of motion: For every action, there is an equal and opposite reaction.
Law of conservation of momentum: It states that the total momentum of a system of objects remains constant in the absence of any external force.
Friction: When a body slides or rolls on the surface of another body, then the force which opposes the motion of the body is called friction

Read and Learn More KSEEB Solutions for Class 9 Science 

Class 9 Social ScienceClass 9 ScienceClass 9 Maths

 

KSEEB Solutions For Class 9 Science Chapter 9 Force And Laws Of Motion

Force And Laws Of Motion Exercises

Question 1. Which of the following has more inertia?
1) a rubber ball and a stone of same size
2)a bicycle and a train
3) a five – rupees coin and a one-rupee coin?
Answer:
1) stone
2) train
3) a five rupees coin

Question 2. In the following example, try to identify the number of times the velocity of the ball changes: A football player kicks a football to another player of his team who kicks the football toward the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also, identify the agent supplying the force in each case.
Answer:
The velocity of the ball changes 3 times.
In the 1st case, a football player of one team supplies the force In the 2nd case, another player of the same team supplies the force In the 3rd case, the goalkeeper of the opposite team supplies the force

Question 3. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer: It is possible for an object to have a non-zero velocity when the ne force is zero. For example, when raindrops fall with a constant velocity, their weight is balanced by the viscous force of the air. Thus the net force will be zero. The magnitude and directions remain unchanged.

Question 4. A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(1) the batsman did not hit the ball hard enough.
(2) velocity is proportional to the force exerted on the ball.
(3) there is a force on the ball opposing the motion.
(4) there is no unbalanced force on the ball, so the ball would want to come to rest.
Answer: (3) there is a force on the ball opposing the motion.

Force And Laws Of Motion KSEEB Class 9 Question Answers 

Question 5. A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400m in the 20s. Find its acceleration and force acting on it if its mass is 7 tonnes (1 tonne = 1000kg)
Answer:
Given, initial velocity (u) = 0, distance travelled (s) = 400m and t = 20s
Using the equation s = ut + 1/2  a
t2,
we get 400 = 0 +1/2 a
(20)2=> a= 2 x 400 /400

= 2ms-2

Force (F) acting the truck= ma = 7000 x 2

– 14, 000N

Question 6. A stone 1 kg is thrown with a velocity of 20m/s across the frozen surface of a lake and comes to rest after traveling a distance of 50m. What is the force of friction between the stone and the ice?
Answer:
Given data,
Mass (m) = 1kg, initial velocity (u) = 20m/s, final
velocity(v) = 0 and distance travelled (s) = 50m
Using the equation   v2 – u2 = 2as, we get
0 – (20)2 – 2a x 50 which gives a = – 400 /100
= – 4 m/s2
The negative sign indicates the retardation of motion of the body . The force of friction between the stone and ice
(F) = ma
= 1 x (- 4) = – 4N

Question 7. An 8000 kg engine pulls a train of 5 wagons, each of 2000kg along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction of5000N, then calculate
(1) the net accelerating force
(2) the acceleration of the train and
(3) the force of wagon 1 on wagon 2
Answer:
(1) The net accelerating force
= Engine force – frictional force

= 40000 – 5000 = 35000N

(2) The acceleration of the train = a = F/m [ Since F = ma]

m = mass of the train = 8000 + 5 x 2000 =18000

Hence a = 35000/18000 = 1.94 m/s2

(3) The force wagon 1 on wagon 2 = The net accelerating force – a mass of

wagon 1 x acceleration of the train = 35000-2000 x 1.94 = 31.111 N

Question 8. An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and the road if the vehicle is stopped with a negative acceleration of 1.7 m/s2?
Answer:
Given, mass (m) = 1500 kg,

acceleration (a) = 1.7m/s2

The force required to stop the vehicle =ma

= 1500 x – 1.7 — – 2550N

Question 9. What is the momentum of the object of mass m, moving with a velocity v?
(1)(mv)2
(2)Mv2
(3) 1/2 mr
(4) mv
Answer: (4)  mv

Question 10. Using a horizontal force of 200, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer: For the constant velocity, the net force on the wooden cabinet must be zero. This means that the friction force must be equal to the applied force ie 200N.

Question 11. Two objects, each of mass 1.5kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5m/s before collision during which they stick together. What will be the velocity of the combined object after collision?
Answer:
Given,
m1= m2 = 1.5kg and u1=2.5m/s  , u2 =- 2.5m/s
According to the law of conservation of momentum
Total momenta after collision = Total momentum before the collision

\(\left(m_1+m_2\right) v=m_1 u_1+m_2 u_2\)

(1.5 + 1.5) v = 1.5 2.5 + 1.5 x (- 2.5)

3v = 3.75 -3.75 = 0
v=0

Class 9 Science Chapter 9 KSEEB Textbook Solutions 

Question 12. According to the 3rd law of motion, when we push on an object, the object pushes back on us with an equal and opposite force. If the an object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer: The student’s justification is wrong because action and reaction act on two different bodies and cannot cancel each other. The exact reason is that the frictional force between the tires and the road is greater than the push on it.

Question 13. A hockey ball of mass 200g traveling at 10m/s is struck by a hockey stick so as to return it along the original path with a velocity of 5m/s. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick
Answer:
Given data,

Mass(m) = 200g = 0.2kg, Initial velocity (u) = 10m/s and final velocity (v) – 5m/s.
According to Newton’s 2nd law of motion, Change in the momentum of the hockey ball = m( v – u)

= 0.2 (- 5 – 10)

= – 3 kgm/s

Question 14. A bullet of the mass 10g traveling horizontally with a velocity of 150m/s strikes a stationary wooden block and comes to rest in 0.03 seconds. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
Given data, mass (m) = 10g = 0.01kg, initial
velocity (u) = 1 Om/s
Final velocity (v ) =  0 and time (t) = 0.03 s.

Using the equation, v = u + at, we get
a = (v – u)/t – (0 – 10)/0.03 = – 500Om/s2 (retardation)

Distance of penetration (s) = ut + 1/2 at2= 10
(0.03) + l/2 (-5000) (0.03)2 = 2.25 m

The magnitude of the force (F) = ma

= 0.01 x (5000) = 500N

Question 15. An object of mass 1 kg traveling in a straight line with a velocity of 10m/s collides with and sticks to a stationary wooden block of mass 5kg. Then they both off together in the same direction. Calculate the total mo men turning just before and after the impact. Also, calculate the velocity of the combined object
Answer:
Let m1 = 1 kg, m2= 5kg, U1= 10 m/s and u2 = 0

Total momentum just before the impact = m1 u1+m2 u2

= 1(10) + 5 (0) = 10kgm/s

According to the law of conservation of momentum, total momentum just after the impact = total momentum just before the impact = 10kgm/s

Hence(m1 + m2 )v = 10 where v = velocity of the combined object.

ie (1 + 5) v = 10 => v= 10/6 = 5/3 = 1.67 m/s

Question 16. An object of mass 100kg is accelerated uniformly from a velocity of 5m/s to 8m/s in 6 1 . Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object
Answer:
Let m= 100kg, u = 5m/s and v = 8m/s

Initial momentum=mu = 100 (5) = 500 kgm/s
Final momentum=mv= 100 (8) = 800 kgm/s

According to Newton’s 2nd law of motion, F =Change in momentum/time

= (800-500) /6 = 50N

Question 17. For a statement refer textbook.
Answer: Raghul was correct ie both the motor car and the insect experience the same force and the same change in momentum. But the impact is greater on the
insect due to its small mass. Hence the insect died.

Question 18. How much momentum will a dumbbell of mass 10kg transfer to the floor it falls from a height of 80cm? (Take downward acceleration = 10m/s2)
Answer:
Given u = 0, s = 80cm = 0.8m and a = 10m/
s2

We have v2 – u2 = 2as => v2-0 = 2 x 10 x 0.8

= 16 => v2= 16 ie v – 4m/s.

Hence momentum = mv = 10 x 4 40kgm/s

Force And Laws Of Motion Additional Questions

One Mark Questions

Question 1. Name the force which is responsible for a change in the position or state of an object.
Answer: Unbalanced forces

Question 2. How is inertia related to mass?
Answer: Inertia is directly proportional to the mass

Question 3. When we shake a fruit tree, fruits fall from the tree. Name the property
Answer: Inertia of rest

Question 4. Three balls made up of aluminum, steel and wood have the same volume and shape. Which of these has the highest inertia? Why?
Answer: Steel ball has the highest inertia because it has a maximum mass

KSEEB Solutions For Force and Laws Of Motion Short Notes 

Question 5. Name the forces acting on a book kept on the table.
Answer: Gravitational force and frictional force

Question 6. How is inertia related to mass?
Answer: Larger the mass, the greater will be the inertia of the body.

Question 7. Mention S.I. unit of momentum kilogram -meter per second (kg ms1)
Answer: Kilogram – Metre per second (kg ms-1)

Question 8. Define the recoil velocity of a sun.
Answer: The velocity with which the gun moves back-ward when a bullet is fired horn it is forward direction is called the recoil velocity of the sun.

Question 9. Name the scientist who proved that objects need no net force for uniform motion.
Answer: Galileo

Question 10. What is the product mass of a body and its velocity called?
Answer: Momentum of the body.

Question 11. Name the physical quantity whose S.I. unit is lOg ms1
Answer: The momentum of a body

Question 12. How is force related to acceleration?
Answer: Acceleration is proportional to the magnitude of the force.

Question 13. Name the physical quantity which correct seconds to a rate of change of momentum.
Answer: Force

Question 14. On what principle a rocket work?
Answer: Law of conservation of momentum.

Question 15. Name the physical quantity which is a measure of inertia
Answer: Mass

Question 16. Name the scientist who gave the famous three laws of motion
Answer: Newton

Question 17. Name the forces acting on a book kept on the table.
Answer: Gravitational force and frictional force

Question 18. How is inertia related to mass?
Answer: Larger the mass, the greater will be the inertia of the body.

Question 19. Mention S.I. unit of momentum kilogram – meter per second (kg ms1)
Answer: Kilogram – Metre per second (kg ms-1)

Question 20. Define the recoil velocity of a sun.
Answer: The velocity with which the gun moves back-ward when a bullet is fired from it is forward direction is called the recoil velocity of the gun.

Question 21. State Newtont’s first law of motion
Answer: A body at rest or in uniform motion will remain so unless an unbalanced force acts on it

Question 22. Name the force which keeps a body moving in a circular path with constant acceleration
Answer: Centripetal force

Question 23. Define the momentum of a body
Answer: The amount of motion possessed by a body during its motion is called the momentum of the body

Question 24. How is force related to an acceleration of a body?
Answer: Force is directly proportional to acceleration

Question 25. Name the quantity measured by he product of mass and velocity
Answer: Momentum=mv

Question 26. On what principle a rocket work?
Answer: The law of conservation of momentum

Question 27. Name the quantity which corresponds to rate of change of momentum.
Answer: Force

Question 28. State Newton’s third law of motion
Answer: To every action, there is an equal and opposite reaction     

Question 29. State the law of conservation of momentum
Answer: The total momentum of a system of objects in the absence of any external force remains constant.

Force and Laws Of Motion Class 9 KSEEB MCQ Solutions 

Question 30. Define friction
Answer: When a body slides over the surface of another force, a force comes to act against the motion. This force is called friction

Question 31. Name the S.I units of
1) force
2)momentum

Answer:
1)S. I unit of force is newton
2)S. I unit of momentum is kgm/s

Question 32. Name the force which slows down a moving vehicle when we apply break.
Answer: Frictional force

Question 33. Name any two effects of a force on a body
Answer:
1) change in speed
2) change in direction of motion
3) change of shape

Question 34. What happens when a force is applied to a rigid body?
Answer: The body undergoes changes in shape and size

Question 35. A ball of mass 20g is moving with a velocity of 40m/s. What is its momentum?
Answer: Momentum = mv = 0.02kg x 40 m/s = 8 kg m/s

Question 36. What force is required to produce an acceleration of 5m/s2 in a body of mass 10kg?
Answer: F = ma = 10 5 = 50N

Question 37. A force of 10 N is acting on a body of mass 2kg. Calculate acceleration produced in the body
Answer: Acceleration (a) = F/m =10/2 = 5 m/s2

Question 38. Which would require greater force – accelerating 1kg mass at 10m/s2 or a kg mass at 4m/s2
Answer: F1 = 1 x 10 = 10N and F2 = 2 x 4 = 8 N

Question 39. Name the property of which a body resists a change in its state of motion. Name the physical quantity which is a measure of this property
Answer: Inertia and mass

Question 40. A man stands on frictionless ice in the middle of the pond. If he wants to reach the shore, what should be done?
Answer: He has waved his hands in the direction opposite to the shore

Question 41. What is the momentum of a toy car of mass 500g moving with a velocity of 2 ms-1
Answer: P mv = 500g x 2 ms-1

= 1/2 kg x 2 ms 1 = 1 kg ms-1

Question 42. Define the term force
Answer: A force is defined as a pull or push which tends to change the state of rest or uniform motion or direction of the motion.

Two Marks Questions

Question 43. Define resultant force
Answer: A single force that produces the same effect as that of a number of forces acting on a body is called the resultant force change in its state of rest or uniform motion in a straight line is called inertia of the body.

Question 44. State Newton’s second law of motion
Answer: The rate of change of momentum of a body is directly proportional to the force and the change takes place in the direction of the force

Question 45. Define inertia of rest. Give an example
Answer:  The natural tendency of a body to remain in its state of rest is called inertia of rest. Example: When we shake the we clothes, water drops come out of the cloth due to the inertia of rest

Question 46. A bullet of mass 20 was horizontally fired with a velocity of 150ms-1 from a pistol of mass 2kg. What is the recoil velocity of the pistol?
Answer:
Net momentum before firing =Net momentum after firing
(2 + 0.02) x 0 = (0.02 x 150) + 2 x v 0 = 3+ 2v => 2v = – 3

v= -3/2  = -1.5ms-1

Negative sign indicates the gun moves in the di¬rection of motion of bullet.

Question 47. A car and a truck have the same momentum. Whose velocity is more and why?
Answer:
The velocity of the car is more.

Given   P1 = p2
Mc Vc = Me Ve
Where    Mc = Mass of car
Vc   = Velocity of car

\(  \mathrm{M}_{\mathrm{t}}\) = Mass of truck

Ve    = Velocity of truck

Now    \(\frac{M_E}{M_C}=\frac{V_c}{V_E}\)

[/latex]M_E>M_C \Rightarrow \frac{M_E}{M_C}>1[/latex]

Question 48. Mention one advantage and disadvantage of friction.
Answer: Friction plays an important role in breaking the system of vehicles. The wearing and tearing of tires is due to friction between tires and the road.

Question 49. Define inertia of motion. Give an example
Answer: The natural tendency of a body to remain in its state of uniform motion in a straight line is called inertia of motion
Example: When a bus driver applies brakes suddenly to stop a moving bus, the passengers in the bus move forward due to inertia of motion.

Question 50. Define inertia of direction. Give an example
Answer: The inability of a body to the change of direction of the motion itself is called inertia of direction Example: When a knife is sharpened by a grinding stone, the streaks of fire appear tangential to the stone. This is due to the inertia of direction.
Example: When a car suddenly turns around a circular path, the passengers in the car are thrown outwards due to inertia of direction.

Question 51. What are the changes possible on an object at rest when
(1) a balanced force
(2) an unbalanced force act on It?
Answer:
1) The balanced force changes the shape and size of the object
2) The unbalanced force changes the state of rest or uniform motion of the body.

KSEEB Class 9 Science Chapter 9 Important Questions 

Question 52. Differentiate between balanced and unbalanced forces
Answer:
Balanced forces: The forces whose resultant is zero are called balanced forces. They may change the shape of the body.

Unbalanced forces: The forces whose resultant is not zero. They change the state of rest or uniform motion of a body.

Question 53. What happens if a fielder stops the fast-moving ball suddenly? Justify your answer.
Answer: The fielder gets his hand hurt by stopping the ball suddenly. Due to the short time of impact, the rate of decrease of momentum of the ball is large and it exerts a large force on the hands of the fielder which may hurt his hands.

Question 54. The following is the distance-time table of an object in motion :

Force and Laws of Motion

(1) What conclusion can you draw about acceleration? is it constant, increasing, decreasing, or zero?
(2) What do you infer about the forces acting on the object?
Answer: Using the formula  \(\mathrm{V}=\frac{s_2-s_1}{t_2-t_1}\)
and  \(a=\frac{v_2-v_1}{t_2-t_1}\)

we get a = 1 m/s2, a= 6 m/s2 ,a3 = 12 m/s2 ,a4 = 18 m/ s2, a5= 24 m/ s2, a6 = 30 m/s2 and a7 = 36 m/s2. These values indicate
(1) acceleration increases un informally
(2) Since F x acceleration, the force also increases uniformly.

Question 55. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort).
Answer: The force applied by 3rd person=ma = 1200 x = 0.2 = 240N

Since all three persons use the same muscular effort, each person exerts a force 250 N.

Question 56.A hammer of mass 500g. moving at 50 ms1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Answer: For the hammer, mass(m) = 500g = 0.5kg, u = 50m/s, t = 0.01s and v – 0 The force of hammer on the nail = ma = m(v-u)/t              ;

= 0.5 (0 – 50) / 0.01 = – 2500N Since the nail stops the hammer, the force of nail on the hammer =- F = 2500N     •

Three Marks Questions

Question 57.
1) State Newton’s third law of motion
2) Express Newton’s second law mathematically explaining the symbols used
3) Define the SI unit of force from this expression
Answer:
1) The rate of change of momentum of a body is directly proportional to the force and the change takes place in the direction of the force
2) Mathematical expression for the force is F = ma where m = mass and a = acceleration of the body
3) When m = 1 kg and a = 1 m/s2 , then F = 1 newton. Thus one newton is the force that produces unit acceleration in the unit mass.

Question 58. State and explain Newton’s third law of motion
Answer:
Statement: To every action, there is an equal and opposite reaction

Explanation:

Chapter 9 Newton's Third Law of Motion

Let us consider two spring balances connected together as shown in the above fig. The fixed end of balance B is attached with a rigid support, like a wall. When a force is applied through the free end of spring balance A. it is observed that both the spring balances show the same readings on their scales. It means that the force exerted by spring balance A on balance B is equal but opposite in direction to the force exerted by the balance B on balance A, The force that balance A exerts on balance B is called the action and the force of balance B on balance A is called the reaction. This gives us an alternative statement of the third law of motion i.e., to every action there is an equal and opposite reaction. However, it must be remembered that the action and reaction always act on two different objects.

Question 59. Explain the working of a rocket based on the law of conservation of momentum
Answer: When a rocket is fired, a large of hot gases due to the combustion of fuel is thrown out vertically downward with a large velocity and hence a large momentum toward the ground. According to the law of conservation of momentum, the rocket gets an equal and opposite momentum due to which it moves vertically upwards.

Question 60. Give an example
1) How force can change the velocity of a body
2) balanced forces
Answer:
1) When we apply brakes to a moving car, the
car slows down
2) In a tug of war, the rope does not move in any direction due to equal and opposite forces (balanced forces) applied by the two teams

Question 61. Give reason:
1) When we step on the peel of a banana, we slip
2) Athletes in a high jump event made to fall either on a cushioned bed or on a sand bed
3) The tires of the vehicle wear out
Answer:
1) The smooth surface of the peel reduces the friction
between the feet and the peel. Hence we slip
2) When the athlete falls on a cushioned bed or a sand bed, the time interval for which the force acts increases, and the rate of change of momentum and hence the force of reaction decreases. Thus the athlete does not hurt.

Newton’s Laws Of Motion Explained Class 9 KSEEB Solutions

Question 62. Give reason:
1) Shock absorbers are used in cars and motorcycles
2) glass or chinaware is packed with straw
3) road accidents at high speeds cause more damage than accidents at low speeds
Answer:
1) The shock absorbers increase the time interval
of the jerk which in turn decreases the rate of change of momentum and reaction of the force on the passengers
2) The straw reduces the time interval of jerk during transportation. this reduces the effect of the force between chinawares and avoids damage.
3) At high speeds, the time of impact of vehicles is very short. So they exert very large forces on each other. Hence damage is more.

Question 63. Give reason:
1) When a gun is fired, the gun recoils
2) As the sailor jumps in the forward direction, the boat moves backward
3) We can’t walk with legs perpendicular to the ground
Answer:
1) When the gun is fired, it exerts a forward force on the bullet. In turn, the bullets exert an
equal and opposite force (Newton’s 3,d law) on the gun. Hence the gun recoils
2) As the sailor jumps in a forward direction(action), he exerts equal and opposite force on the boat. Hence the boat moves backward.
3) We our legs are perpendicular to the ground, the .ground exerts an equal and vertically opposite force on the legs. Hence we feel motionless.

Question 64. Give reason:
1) It is easier to stop a tennis ball than a cricket ball moving with the same speed
2) All cars are provided with seat belts
3) Water sprinkler used for grass lawns begins to rotate as soon as the water is supplied
Answer:
1) The mass of a tennis ball is smaller and hence it has smaller momentum and a smaller force is required to stop it.
2
) Seat belt increases the time interval of forward motion during the accident. This reduces the reactionary force on the passenger and prevents serious injury
3) The rushing water out of the sprinkler exerts a reactionary force on the sprinkler.

Question 65. From the air-filled balloon, the air is released from its mouth in a downward direction. Write the other observations made by you and justify your answer.
Answer: The balloon moves vertically upward direction due to Newton’s 3rd law of motion.

Question 66.
1) Define momentum.
2) A ball weighing 500g is thrown vertically upward with a speed of lOm/s. What will be its momentum

  • initially
  • at the highest point?

Answer:
1) Refer ans of Q. 11

Given mass(m) = 500g = 0.5kg andu= 10m/s

Initial momentum = mu = 0.5 x 10 = 5N

The highest point, v = 0. Hence momentum

= 0.5 x 0 = 0

Four Marks Questions

Question 67.
1) State the law of conservation of momentum.
2) A body of mass 2 kg initially moving with a velocity of lOm/s, collide with another body of mass 5kg at rest. After a collision, the velocity of the first body becomes Im/s. Find the velocity of the second body.
Answer:
1) refer ans of Q. 18
2) Before collision: m1= 2kg, m2 = 5kg, u1 = 10m/s and u2 = 0

After a collision, v1 = 1m/s, and v2 =?

According to the law of conservation of momentum,

Final momentum= Initial momentum

\(\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2=\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2\)

2(10) + 5(0) = 2(1) + 5(V2)   =>  5v2 + 2 = 20 =>

5v2 = 18

Hence v2 = 18/5 = 3.6m/s .

Question 68.
1) State Newton’s first law of motion.
2) Deduce Newton’s first from Newton’s second law of motion
3) What will be the acceleration of the car of mass 1200 kg on applying a force of 120N on it?
Answer:
1) Refer Ans of Q. 9
2) According to Newton’s 2nd law of motion, F = ma = m(v – u) /t where u = initial velocity, v = final velocity and m = mass of the body

If F = 0, then m(v – u) = 0 => v u = 0 (since m=O)

Thus we get v=u ie uniform motion

Thus a body is confined with uniform motion when the force on it is zero.

This is the statement of newton’s first law of motion

3) acceleration of the car = Force/mass = 120 /1200 = 1/10-0. lm/s2

KSEEB Chapter 9 Class 9 Detailed Solutions On Force 

Question 69.
1) State Newton’s second law of motion.
2) Derive the expression for the force in terms of acceleration and mass of a body
3) A cricket player lowers his hands while catching a ball. Why?
Answer:
1) Refer ans of Q. 13
2) Consider an object of mass ‘m’ moving with an initial velocity ‘u’ along a straight line. It is accelerated uniformly to a velocity ‘v’ by applying a constant force ‘F’.

Initial momentum of the object ( p1 ) = mu

Final momentum of the object (p2) = mv

The change in momentum=p2– p1 = mv – mu

= m(v-u)

The rate of change of momentum

= m(v-u)/t

According to Newton’s second law of motion F cc m (v – u) /t = k ma

Question 70.
1) State the law of conservation of momentum
2) Prove the law of conservation of momentum by taking the case of two balls
3) A bullet of mass 20g is horizontally fired with a velocity of 150m/s from a pistol of mass 2kg. What is the recoil velocity of the pistol?
Answer:
1) Refer ans of Q. 18.
2) Consider two ball A and B of masses mA and mB are traveling in the same direction along a straight line at different velocities uA and (uA > uB) respectively.

Let the balls collide each other as shown in the figure. During the collision, ball A exerts a force FABon ball B and ball B exerts a force FDA on ball A.

Let vA and vB be the velocities of the two balls A and B respectively after the collision. If he t be the time of impact of the collision, then

The rate of change of momentum of ball A=  \( m_A\left(v_A-u_A\right) / t\)

The rate of change of momentum of ball B =\(\left(v_B-u_B\right) / t\)

According to Newton’s 3rd law of motion

\(\mathrm{F}_{\mathrm{AB}}=-\mathrm{F}_{\mathrm{BA}} \Rightarrow \mathrm{m}_{\mathrm{A}}\left(\mathrm{v}_{\mathrm{A}}-\mathrm{u}_{\mathrm{A}}\right) / \mathrm{t}\) \(=-\mathrm{m}_{\mathrm{B}}\left(\mathrm{v}_{\mathrm{B}}-\mathrm{u}_{\mathrm{B}}\right) / \mathrm{t}\)

This gives  \(  \mathrm{m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}} \mathrm{u}_{\mathrm{B}}=\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}\)

ie Total momentum of balls A and B before collision = Total momentum of the balls A and B after the collision . This proves the law of conservation of momentum
3) Total momentum of the pistol and bullet before firing = (2 + 0.02)kg x 0 m/s = 0

The total momentum of the pistol and bullet after the firing – (0.02 x 150) + ( 2 x v) where v – velocity of recoil of the gun

Total momentum after the firing =total momentum before firing 3+2v = 0 => v = – 3/2 = – 1.5m/s

Question 71. Explain Galileo’s experiment with marbles to prove that no net force is needed for uniform motion.
Answer:

Galileo's experiment with marbles Chapter 9

When a marble rolls down an inclined plane, its velocity increase. But its velocity decrease gradu¬ally when it climbs up the opposite inclined plane. This is due to friction which opposes the mo¬tion. However when the inclined planes are fric¬tionless, the marble roll down the slope of left inclined plane and moves up the right and opposite inclined plane to the same height from it which it was released.

If the angle of inclination of the right side plane is gradually decreased, the marble travels more and more distances in order to reach its original height. If the right side plane is made completely horizontal, the marble would continue to move to reach the same height from which it was released. This means that no net force is required for uniform motion.

Question 72.
1) Define force. Mention S.I. unit of force
2) Mention any two possible effects of force
Answer:
1) Force is defined as a push or a pull that changes or tends to change the state of rest or uniform motion or direction of motion of a body.
S.I, unit force is Newton,
2) The possible effects of force are

  • change the speed of the body
  • change the direction of the motion
  •  change the shape and size of the body

Question 73.
1) State Newton’s three laws of motion
2) Define the inertia of a body
Answer:
First law: A body at rest or in uniform motion will remain so unless an unbalanced force acts on it.
Second law: The rate of change of momentum of body is directly proportional to the applied unbalanced force and the change takes place in the direction of the force.
Third law: Action and reaction are equal and opposite and they act on different bodies.
Inertia: The natural tendency of a body to oppose any change in its state of rest or uniform motion in a straight line is called the inertia of the body.

Force And Laws Of Motion KSEEB Class 9 Question Answers 

Question 74. State Newton’s first law of motion. Explain it with the help of suitable examples.
Answer:
Newton’s first law:
A body at rest or in uniform motion will remain so unless an unbalanced force acts on it.

Explanation: According to Newton’s first law, a body at rest continues to be rest. For example, a passage sitting in a bus moves backward, when the bus starts suddenly. When the bus moves forward, his foot which is in contact with the moves forward, while his upper body moves backward.

The law also states that a body in uniform motion continues to be in the straight path. This can be explained in the following example.

When a mining bus suddenly stops by applying brakes, a passenger standing in the bus moves forward. When the bus stops suddenly, the lower part of the body comes to rest while the upper part tends to move forward.

Thirdly the law states that a body with uniform motion along a straight path cannot change its di¬rection itself When a moving bus turns around a sharp bend, a passenger in the bus tends to move sidewise. His lower part moves in the direction of motion while his upper part tends to be in the original direction.

Question 75.
1) State and explain Newton’s second law of motion.
2) A cricket player lowers his hands while catching a fast-moving ball. Explain.
Answer:
Newton’s Second Law
1) The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force. When a larger force is applied on a body, the velocity of the changes, and hence momentum = mv also changes. This change in momentum is proportional to the force.

2) By lowering their hands, the fielder increases the time during which the high velocity of the moving ball decreases to zero. Thus acceleration of the ball is decreased and therefore the impact of catching the fast-moving ball is also reduced. If the ball is stopped suddenly, then its high veloc¬ity decrease to zero in a very short interval of time. Thus the rate of change of momentum of the ball will be very large. Therefore, a larger force would have to be applied for holding the catch that may hurt the palm of the fielder.

Question 76.
1) State and explain Newton’s second law of motion.
2) A force of 5N gives a mass m1. an acceleration of 10ms’2 and a mass m2, an acceleration of 20ms”2. What will be acceleration produced by the same force if both the masses were tied together?
Answer:
1) Refer the above answer
2) From Newton’s 2nd law of motion

\(m_1=F / a_1=5 / 10=0.5 \mathrm{~kg}\) \(m_2=F / a_2=5 / 20=0.25 \mathrm{~kg}\)

Now combined mass = 6.5 + 0.25 = 0.75kg acceleration of combined

Mass (a) = \(\frac{F}{m_1+m_2}=\frac{5}{0.75}=6.67 \mathrm{~ms}^{-2}\)

Question 77. A constant force acts on an object of mass 5kg for a duration of 25. It increases the object’s velocity from 3ms-1 to 7ms-1. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5S what would be the final velocity?
Answer:
Case (1): Given

Initial velocity (u) – 3ms-1
Final velocity (v) = 7 ms-1
mass (m) = 5kg
time (t) = 2s

According to Newton’s 2nd law of motion F = ma

F = m(v – u)/t

F = \( \frac{5(7-3)}{2}=\frac{5 \times 4^2}{\not 2}\)

F = 10N

case (2): When t — 5s    than 10 = 5(v – 3)/5

50 = 5 (v – 3) => 10 = v- 3 => v= 10 + 3 => v = 13 ms-1

Question 78.
1) State and explain Newton’s third law of motion.
2) What do you mean by the recoil of seen action and reaction are equal and opposite and they act on different bodies?
Answer:
1)When one object exerts a force (action) on another object than the second object also exerts a force (reaction) on the first. These two forces are always equal in magnitude but opposite in
direction. The third law indicates that a single force can never exist.
2) When a bullet is fired, the gun exerts a for-ward force on the bullet. In turn the bullet exerts an equal and opposite reaction force on the gun. As a result, the gun moves backward. This is known as the recoil of the gun.

Question 79. Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60kg and was moving with a velocity 5ms’1 while other has a mass of 55kg was moving faster with a velocity 6ms1 toward the first player. In which direction and with what velocity will they move after they become entangled? Assume the frictional force acting on them is negligible.
Answer:
Let m1 = 60kg     u1= 5 ms-1

m2 = 55kg      u2 = -6ms-1

Before collision,
Net momentum=  \(m_1 u_1+m_1 u_1\)

= 60 x 5 + 55 x (-6)

= 300 – 300 = -30 kg ms-1

After collision
Net momentum = ( m1 + m2 ) V = (60 + 55)V   = 115 V

According to the law of conservation of momentum

Net momentum before the collision = Net momentum after the collision

– 30 = 115 V => V = (-30/115) = -0.26 ms-1

The entangled players move in the direction the second player was moving before the collision.

Force And Laws Of Motion  Application Questions

Question 1. Name an art based on Newton’s third law of motion.
Answer: Karate

Question 2. Which law of motion is called the law of iner¬tia?
Answer: Newton’s first law of motion.

Question 3. Name the unbalanced force which slows down a moving bicycle when we stop pedaling it.
Answer: Friction

Question 4. A person walks West to East direction. What will be the direction of friction between his feet and road.
Answer: East to west

Question 5. We tend to fall when we step on the peel of a banana. Why?
Answer: Peel of bananas reduces the friction between our feet and the road. Due to unbalanced force, we tend to fall.

Question 6. Explain why some of the leaves may be detached from a tree if we vigorously shake its branch.
Answer: Initially, the leaves are at rest. When a branch is shaken vigorously, leaves are set into motion. This motion is opposed by the inertia of the rest of the leaves. Hence they get detached.

Question 7
1) Why do the passengers in a bus tend to fall forward when it suddenly stops?
2) Why do the passengers in a bus tend to fall back when it starts suddenly?
Answer:
1) When the bus comes to rest suddenly, the lower part of the body which in contact with the floor of the bus also comes to rest. At the same time, the upper part of the body has the tendency of moving due to the inertia of motion. Hence the passenger fells forward.
2) When a bus accelerates suddenly, the lower part of the body which is in contact with the floor of the bus is set into motion. At the same time, the upper part of the body has a tendency of rest due to the inertia of rest. Hence the passenger falls backward.

Question 8. When a carpet is beaten with a stick, dust comes out of it. Explain why?
Answer: Initially, the dust is at rest on the carpet. When the carpet is beaten with a stick, the dust is set into motion. But this motion is opposed by the inertia of the rest. As a result, the dust gets detached from the carpet.

Question 9. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer: When the bus starts suddenly or stops suddenly or takes a turn, the luggage may fall due to inertia of rest or motion or direction kg-m/s-1

KSEEB Solutions For Class 9 Science Chapter 11 Work And Energy

KSEEB Solutions For Class 9 Science Chapter 11 Work And Energy Important Concepts

Work: Work is said to be done if a body is displaced by a force. Mathematically, W= Force x displacement. Work is a scalar quantity.
S.I unit of work: newton-meter or joule
Positive and Negative work: If the angle between the directions of displacement and force is an acute angle(< 90°), then the work is said to be positive. If the angle between the directions of displacement and force is an obtuse angle (> 90), then the work is said to be a negative obtuse angle (> 90), then the work is said to be negative.
Zero work: Work is said to be zero if
1) The angle between the directions of displacement and force is 90° or displace is zero or no force acts on the body
Energy: Energy is defined as the capacity of a body to do work. It is a scalar quantity and its S.I unit is the joule.

Read and Learn More KSEEB Solutions for Class 9 Science 

Types of energy: Mechanical energy (kinetic energy and potential energy), heat energy, chemical energy, electrical energy, nuclear energy, light energy etc.
Kinetic energy: It is the energy of a body by the virtue of its motion
Potential energy: It is the energy possessed by a body by virtue of its position or configuration
Law of conservation of energy: Energy can neither be created nor destroyed
Power: Power is defined as the rate of doing work. It is a scalar quantity and the S.I unit is a watt.
Commercial unit of power: Commercial unit of power is a kilowatt hour (kWh). It is electrical energy used by a device of power 1000W in one hour, kilowatt hour (kWh) 1 kWh = 3.6 x 106 J

Class 9 Social ScienceClass 9 ScienceClass 9 Maths

 

Work And Energy Exercises

Question 1. A force of 7N acts on an object. The displacement is 8m along the direction of a force. What is work done?
Answer: Work = Force x displacement = 7 x 8 = 56 J

Question 2. When do we say that work is done?
Answer: Work is said to be done when a force displaces a body.

Question 3. Write an expression for the work done when a force is acting on an object in the direction of its displacement
Answer: Work = Force x displacement

Question 4. Define 1J of work
Answer: One joule is defined as work done by a force of IN in displacing a body through lm in its direction

Work And Energy KSEEB Class 9 Question Answers 

Question 5. A pair of bullocks exerts a force of HON on a plough. The field being ploughed is I5m long. How much work is done in ploughing the length of the field.
Answer: Work done = Force x displacement = 140 x 15 = 2100J

Question 6. What is the kinetic energy of an object?
Answer: The energy of an object by virtue of its motion is called kinetic energy.

Question 7. Write an expression for the kinetic energy of an object
Answer: K.E = 1/mv2 where m = mass of the object and v = velocity of the object

Question 8. The kinetic energy of an object of mass m moving with a velocity of 5m/s is 25J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased 3 times?
Answer:
Initial kinetic energy=25J Since K.E is proportional to (velocity)2 when velocity is doubled, then
K.E =(2)2x Initial K.E = 4 x 100 = 400J
When the velocity is trebled, then K.E =(3)2x 25 = 225J.

Question 9. What is power?
Answer: Power is defined as the rate of doing work

Question 10.Define 1 watt of power
Answer: One watt is the power when the one-joule work is done in one second

Question 11. A lamp consumes 1000J of electrical energy in 10s. What is its power?
Answer: Power of he lamp = energy consumed / time = 1000/10= 100W

KSEEB Solutions For Class 9 Science Chapter 11 Work And Energy

Work And Energy Textual Questions

Question 1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
1)Suma is swimming in a pond.
2)A donkey is carrying a load on its back.
3)A windmill is lifting water from a well.
4)A green plant is carrying out photosynthesis.
5)An engine is pulling a train.
6)Food grains are getting dried in the sun.
7)A sailboat is moving due to wind energy.
Answer:
1)Yes. Work is done by suma become she is moving forward by pushing water backward
2)No. Work is not done by the donkey. This is because the weight (Force) of the load acts perpendicular to its displacement
3)Yes. Work is done by the windmill by lifting water against the earth’s gravity
4)Yes. Work is not done by the green plant because neither a force nor a displacement is observed during the photosynthesis by leaves
5)Yes. The force exerted by the engine and the displacement of the train are in the same direction
6)No work is done since neither a force nor displacement is seen
7)Yes. Work is done by the wind in pushing the sailboat in the direction of the wind

Question 2. An object was thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer: Zero because the force of gravity and the displacement are perpendicular

Question 3.A battery lights a bulb. Describe the energy changes involved in the process.
Answer: First chemical energy is converted into electrical energy and then electrical energy is converted into heat and light energy.

Question 4. A certain force acting on a 20 kg mass changes its velocity from 5m/s to 2 m/s. Calculate the work done by the force.
Answer:
Work done by the force = Change in the kinetic energy
= Initial kinetic energy – final kinetic energy
= 1/2 mu2– 1/2 mv2= 1/2 m(v2 – u2)
= 1/2 (20) (52-22) = 10 (25 – 4) = 210J

Class 9 Science Chapter 11 KSEEB Textbook Solutions 

Question 5. A mass of 10kg is at point A on a table. It is moved to point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer: Zero because the force of gravity and the displacement are perpendicular

Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer: No. This is because the loss in potential energy is equal to the gain of its kinetic energy such that the total energy of the object remains constant.

Question 7. What are the various energy transformations that occur when you are riding a bicycle?
Answer: Muscular energy is converted into mechanical energy (kinetic energy) of the bicycle and a part of the kinetic energy of the bicycle is converted into heat energy due to friction between the tires and the road.

Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer: Yes. Muscular energy is used to overcome the friction between the rock and the ground

Question 9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer: 1 unit = 1kwh = 3.6 x106J.
250 units = 250 x 3.6 x106J. = 9×106J.

Question 10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is halfway down.
Answer:
Potential energy of the object = mgh
= 40 x 9.8 x 5 = 1960 J
Let v be the velocity of the object a the halfway mark ie when s = 2.5 m
We have  v2= u2+2gs =  02 + 2 x 9.8 x 2.5 = 49 => v = 7m/s
The kinetic energy of the object = 1/2 mv2 = 1/2 x 40 x49 = 980J

Question 11. What is the work done by the force of gravity on a satellite moving around the earth? Justify your answer.
Answer: Zero. Because the force of gravity acts as the dentripetal force which is perpendicular to the displacement.

Question 12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer: The net force acting on the falling raindrops is zero ie rain drops are displaced with no force acting on them.

Question 13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer: No work was done because the displacement of the bundle of hay is zero.

Question 14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer: Energy used = Power x time =1.5 kW x 10 hour =15 kWh

Question 15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer: When we draw a pendulum bob to one side, the work done against gravity is stored as potential energy in the bob. When the bob is released, this potential energy is gradually converted into kinetic energy such that at any position, the sum of kinetic and potential energies is always constant. When the bob is at the mean position, its potential becomes zero and kinetic energy will be maximum. As the bob reaches another extreme position the total energy will only be potential energy. Thus total energy is conserved. Du-friction, the energy of the bob is lost gradually in the form heat and the motion of the pendulum decreases with time and comes to rest after some time.

Question 16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer: Work done = kinetic energy of the object = 1/2 mv2

Question 17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h.
Answer:
Given mass(m) = 1500kg and velocity(v)
= 60km/h = 60 x 5/18 = 50/3 m/s
Work to be done = kintic energy of the car = 1/2 mv2 = 14 x 1500 x (50/3)2= 208,333.33J

Question 18. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive, or zero.

Work And Energy chapter 11Answer:
(1) Since the force(F) is perpendicular to the displacement(s), work done is zero
(2)Since the force(F) and displacement(s) are in the same direction, work done = Fs joule and positive
(3)Since the force(F) and displacement(s) are opposite directions, work done = – Fs joule and negative.

Question 19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer: Yes. I do agree with her. Acceleration can be zero if the resultant of several forces acting on the object is zero.

KSEEB Solutions for Work And Energy Short Notes 

Question 20. Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Answer: The power of each device = 500W = 0.5kW
Therefore, Energy consumed by the 4 devices = 4xPxt = 4x 0.5 x 10 = 20kWh

Question 21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer: On reaching the ground, the kinetic energy of the falling body is converted into heat energy and sound energy.

Work And Energy Additional Questions

One Mark Questions

Question 1. Write an expression for the work when force and displacement are inclined with an angle Q.
Answer:
W = FS cos Q
where F = force
S = displacement

Question 2.Define the term energy
Answer: The energy of a body is defined as the capacity of the body to do the work.

Question 3. Name the largest natural sources of energy
Answer: The sun is the largest natural source of energy.

Question 4. Both work and energy has the same unit joule. Why
Answer: The energy possessed by an object is measured in terms of the capacity of doing work. Hence energy and work have the same unit joule.

Question 5. Give an account for the kinetic energy of a flying arrow.
Answer: The work done in stretching the string is stored as P.E. in the string. When the string is released the P.E. is converted into K.E. of the flying arrow.

Question 6. How do plants produce food?
Answer: Plants produce food through a process called photosynthesis with the help of solar energy.

Question 7. How are fuels such as coal and petroleum formed?
Answer: Millions of years ago plants and animals were buried deep inside the crust of the earth under temperature and pressure. They got converted into coal and petroleum.

Question 8. Calculate the speed of a body of mass 1kg having a kinetic energy of 1J
Answer:  1/2 mv2 = kinetic energy => 1/2  x 1 x v2 = 1 =>v2= 2 and v = \(\sqrt{2} \)m/s

Question 9. What happened to the potential energy and kinetic energy of a stone on reaching at the highest position?
Answer: Potential energy will be maximum and kinetic energy will be zero.

KSEEB Class 9 Science Chapter 11 Important Questions 

Question 10. How much work is done by a weight lifter when he holds a weight of 80kg on his shoulders for 2 minutes?
Answer: zero

Two Marks Questions

Question 11. Identify the energy possed by
(1)A running horse
(2)A raised hammer
(3)Compressed spring
(4)Water stored in a dam
(5)Book kept on a table
(6)A man climbing a hill
(7)A flying bird
Answer:
(1) Kinetic energy
(2) Potential energy
(3) Potential energy
(4) Potential energy
(5) Potential energy
(6) Both kinetic and potential energies
(7) Both K.E and P.E

Question 12. Write the energy transformation at
(1) Thermal power station
(2) Nuclear Power plant
(3) Hydroelectric power station
Answer:
(1) Chemical energy of coal —> heat —> kinetic energy —> electrical energy
(2)nuclear energy —> heat —> mechanical energy —> electrical energy
(3)kinetic energy of water —> electrical energy

Question 13. If the speed of a car is increased 3 times, what is the increase in kinetic energy?
Answer: Since K.E ∞ v2, the kinetic energy of the car increases 9 times

Question 14. A stone is thrown vertically upwards. The kinetic energy of the stone decreases gradually. Is the law of conservation of momentum violated? Justify your answer.
Answer: No. The law of conservation of energy is not violated because the decrease in the kinetic of the stone will be equal to the increase in its potential energy.

Question 15. What will be the power of a body when a force ION moves it with a constant velocity of 2m/s?
Answer: Power = Force x velocity =10 x 2 = 20W

Question 16.Mention the two conditions that need to be satisfied for work to be done
Answer:
1) A force should act on the object
2) The object must be displaced

Question 17. We apply a force to lift an object upwards against the gravity of the earth. Mention which force does positive work and which force does negative work.
Answer: The applied force does positive work since force and displacement are in the same direction. The force of gravity does negative work since the displacement is opposite to the force.

Question 18.
1) How is kinetic energy related to

  • Mass of the body
  • the velocity of the body

Answer:
1) Kinetic energy is directly proportional to the mass of the body.
2) Kinetic energy is directly proportional to the square of velocity.

Question 19. A bullet of mass 5g is fired with a velocity of lOOm/s. What is its kinetic energy?
Answer:
M = 5g = 5 x 10-3kg
v= 100 m/s
K.E. = 1/2 Mv2
= 1/2  x 5  x  10-3  x (100)2 =  25J

KSEEB Solutions Chapter 11 Energy Transformation Class 9 

Question 20. An object of mass 5kg is dropped from a height of 10m. What will be the kinetic energy? When it is 5m from the ground (g = 10m/s)
Answer: W.K.T.
v2=u2 + 2gh
= ( 0 )+ 2 x 10×5
v2= 100
v = \(\sqrt{100}=10 \mathrm{~ms}^{-1}\)

K.E. = 1/2 Mv2= 1/2  x  5  x  100 = 250 J

Question 21. Does a stretched rubber possess energy from where does it set energy?
Answer: Work done to stretch the rubber is converted into the potential energy of the rubber band.

Question 22. What kinds of energy conversions sustain the water cycle?
Answer: Solar energy converts water from water bodies into water varpus which get cooled and pour as rains. When the rainwater flows with K.E. and reaches the water bodies and its K.E. is converted into P.E.

Question 23. Define mechanical energy. Mention two forms of mechanical energy. State the law of conservation of energy
Answer: Energy possessed by a body by virtue of its position of motion is called mechanical energy. The two forms of mechanical energy are potential energy and kinetic energy. The law of conservation of energy states that energy can neither be created nor destroyed

Question 24. Name the following devices which convert
1)Chemical energy into electrical energy
2)Electrical energy into mechanical energy
3)Mechanical energy into electrical energy
4)Light energy into electrical energy
5)Nuclear energy into electrical energy
6)Heat energy into mechanical energy
7)Wind energy into electrical energy
Answer:
1) Electric cell
2) Electric motor
3) Generator
4) Photocell
5) Nuclear power reactor
6) Turbine
7) Wind-mill

Question 25. Define work. When the work is said to be
(1) positive and
(2) negative
Answer:
Work is said to be done if a body moves under the action of a force
(1)If the force and displacement of a body are in the same direction, then work is said to be positive
(2)If the force and displacement of the body are in opposite directions, then work is said to be negative

Question 26. Write an expression for the work when force and displacement are
(1) in the same direction
(2) opposite directions
Answer:
(1) Fs
(2) – Fs where F = force and s = displacement

Question 27. Sharma tried to push a heavy rock of mass 120kg for 2 minutes with a force of 50N but could not move it. What will be the work done by Sharma at the end of 2 minutes?
Answer: Zero because there is no displacement of rock

Three Marks Questions

Question 28. What is kinetic energy? Derive an expression for the kinetic energy of a body
Answer:
Kinetic energy:
The energy of an object by virtue of its motion is called as its kinetic energy. Consider an object of mass ‘m’ moving with uniform velocity ‘u’. Let a constant force F acts on the object and displaces it through a distance ‘s’

Work done by the force W = Fs

Let ‘v’ be the final velocity of the object and ‘a’ be the acceleration of the object.

Using the equation v2=  u2 + 2as, we get

\(\mathrm{s}=\frac{v^2-u^2}{2 a}\)

According to Newton’s 2nd law of motion F = ma

Hence W = mas =ma x ( v2– u2 )/2a                          

\(=1 / 2 m\left(v^2-u^2\right)\)

If the object were at rest before the application of the force, then u = 0, and kinetic energy Ek = 1/2 m(v2 – u2)

Work And Energy KSEEB Class 9 Detailed Solutions 

Question 29.
1) Define gravitational potential energy.
2)Derive an expression for the gravitational potential energy of an object of mass ‘m’ kept a height ‘h from the ground
3)Find the energy possessed by an object of mass 10kg when it is at a height of 6m above the ground (given g = 9.8m/s2)
Answer:
1) The gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity.
2)      Chapter 11 Gravitational potential energy

Consider an object of mass ‘nT is raised through a height ‘h’ from the ground. The force required to do this is F = mg where g = acceleration due to gravity
Thus work done (W) = F x h = mgh
This work is stored as the gravitational potential energy of the object. Thus gravitational potential energy (Ep) = mgh
3) Given m= 10kg, h = 6m and g = 9.8m/ s2 Potential energy of the body = mgh
= 10 x 9.8 x 6 = 588 joule

Question 30.
1) Define power. What is its S.I. unit?
2) Two girls A and B each of weight 400N climb up a rope through a height of 8m. Girl A takes the 20s while B takes the 50s to accomplish this task. What is the power expended by each girl?
Answer: Power is defined as the time rate of doing work.
ie P = work/time.
S.I unit power is a watt
Power extended by the girl A=mgh /1 = 400 x 8/20= 160W
Power extended by the girl B = 400 x 8 / 50 = 64W Self-test

Question 31. Define K.E. of a body. Prove that the change in K.E. of the body is equal to work done on it.
Answer:
K.E. of a body is defined as energy possessed by the body by virtue of its motion.

Consider an object of mass ‘m’ moving with uniform acceleration ‘a’.
u = initial velocity
v = final velocity
s = displacement, then we have v2= u2+ 2as => s =  \( \frac{\left(v^2-u^2\right)}{2 a}\)

\(\mathrm{w}=\frac{1}{2} m\left(v^2-u^2\right)\)

If the object moves from rest, then u = o.
Hence w = 1/2 M V2

Thus work done = change in K.E. of the body.

Question 32.
1)An expression for K.E. of a body.
2)How much work has to be done by the engine of the car to increase the velocity of car from 30kmh-1 to  60kmh-1 Most of car is 1500kg.
Answer:
1)K.E. = 1/2 M V2
Where m = mass of the body v = velocity of the body
2) Mass of the car (m) = 1500kg Initial velocity of the car (u) = 30 x

\(\frac{5}{18}=\frac{25}{3} \mathrm{~ms}^{-1}\) \(1 \mathrm{kmh}^{-1}=\frac{5}{18} \mathrm{~ms}^{-1}\)

Final velocity of the car (v) = \(60 \times \frac{5}{18}=\frac{50}{3} \mathrm{~ms}^{-1}\)

work done by the engine [/latex]=\frac{1}{2} m\left(v^2-u^2\right)[/latex]

\( =1 / 2 \times 1500\left(\frac{50}{3}\right)^2-\left(\frac{25}{3}\right)^2\) \( =750 \times\left(\frac{2500}{9}-\frac{625}{9}\right)\) \(=\frac{750 \times 1875}{9}=156250 \mathrm{~J}\)

 

Work And Energy KSEEB Class 9 Solutions 

Question 33. The gravitational potential energy of a body does not depend on the path on which the body is moved. Illustrate your answer with the example.
Answer:

Chapter 11 Gravitational potential energyCase (1):
consider a block is taken from initial position A to B along the path -1 work done (W1) = Force x displacement
= mg x AB
ie Gravitation P.E. = mg x h

Case (2):
Now consider the block is taken along path 2 ie APQRSB
work done to raise from A to P = mg * AP work done to move from P to Q = 0 (since the displacement is perpendicular to the force of gravity)

work done to move from Q to R = mg x QR

work done to move from R to S=0 work done to move from S to B=mg x SB

thus total work done to raise the block along path 2 = mg (AP + QR + SB)
=mg x AB
Gravitational P.E. = mgh

Thus gravitational potential energy does not de- pend on the path through which the body is raised.

Work And Energy Application Questions

Question 1. A man pushes a wall but fails to displace it. Is work done? Justify your answer.
Answer: No work is done because there is no displacement of the wall.

Question 2. How much work done in moving an object around a circular path of radius 5m by a force of 10N?
Answer: zero. No work is done because the displacement and the centripetal force are perpendicular to each other.

Question 3. A 35kg boy runs along a circular path of radius 10m with uniform speed 5m/s. How much word is done by the in completing one circle. Justify your answer.
Answer: Zero. The displacement of the boy after one circle is zero. Hence work done is zero.

Question 4.
1) When does an object

  • Lose energy
  • gain energy

Answer:
1) When work is done by the object, the body loss energy
2) When work is done on the object, the body gains energy

Question 5. Can kinetic energy be negative?
Answer: because K.E. = 1/2 mv2
where mass (m) = positive quantity
and V2=positive quantity

Question 6. Why does the air move from place to place?
Answer: Due to solar energy the air is heated and becomes light. Hence it moves up and cooler air reaches its place.

KSEEB Solutions For Class 9 Science Chapter 12 Sound

KSEEB Solutions For Class 9 Science Chapter 12 Sound Important Concepts

Sound: It is a form of energy produced by a vibrating body and travels in the form of longitudinal waves
Vibration or oscillation: The to and fro motion of a body about its mean position Amplitude: Maximum displacement of a particle in vibration
Time – period: Time is taken by a particle to complete one vibration
Frequency(f): The number of vibrations executed by a particle of a medium or the number of waves produced by a source. It is measured in hertz (Hz)
Wave: A disturbance set up in a medium due to vibrations of the particles of the medium Wave velocity(v): Distance travelled by the wave in one second
Transverse wave: The wave iu which the particles vibrate perpendicular to the direction of the wave motion.
Examples: Light waves, X- rays, waves on the surface of water etc

Read and Learn More KSEEB Solutions for Class 9 Science 

Class 9 Social ScienceClass 9 ScienceClass 9 Maths

 

Longitudinal wave: The wave in which the particles vibrate along the direction of the wave motion. Examples: Sound waves, waves produced by the vibrating prongs of a tuning fork.
Crest: The maximum displacement of a transverse wave in the positive cycle
Trough: The maximum displacement of a transverse wave in the negative cycle
Compression: The part of the longitudinal wave in which the particles come closer to each other. Density and pressure will be maximum
Rarefaction: It is the part of the longitudinal wave in which particles move apart from their normal positions
Wavelength: For a transverse wave, the wavelength is the distance between two consecutive crests or troughs. For a longitudinal wave, it is the distance between two consecutive compressions or troughs. In general, it is the distance travelled by a wave in one period
Relation between wave velocity, frequency and wavelength: v= f λ
Mechanical waves: The waves which need a material medium for their propagation Example: Sound waves
Electromagnetic waves: The waves which can propagate through a vacuum also Example: Light waves
Echo: The repetition of sound reflection of original sound from the surface of large and hard obstacles. For hearing distinct echoes, the minimum distance of the obstacle from the source of the sound must be 17.2 m.
Law of reflection of sound: The directions in which the sound is incident and reflected make equal angles with the normal to the reflecting surface at the point of incidence and the three lie in the same plane.
Reverberation: The persistence of sound in a hall or an auditorium due to multiple reflections of sound
Musical sound: It is the sound which produces a sensation of pleasure.
Noise: It is the sound which is not pleasing
Pitch: It is a property of sound which enables a person to differentiate between a high and flat sound. Higher is the frequency, the greater will be the pitch of the sound
Loudness: It is a physiological response of the ear to the intensity of the sound
Quality(Timbre): It is the property of the sound which enables a person to differentiate between two sounds of the same pitch and loudness.
Audible sound: It is the sound whose frequency range is 20Hz to 20kHz
Infrasonic waves: The sound waves whose frequency is less than 20Hz
Ultrasonic waves: The soundwaves with frequencies greater than 20kHz
SONAR(Sound Navigation and Ranging): It is a technique used to detect the direction and distance of underwater objects by using ultrasonic sound waves.
Megaphone: A device used to send sound in a particular direction
Stethoscope: A medical instrument used for listening to sounds produced within the body mainly in the heart and lungs.

Sound KSEEB Class 9 Question Answers 

KSEEB Solutions For Class 9 Science Chapter 12 Sound

Sound Exercises

Question 1. How does the sound produce by a vibrating object in a medium reach your ear?
Answer: The vibrating object sets neighbouring air particles into periodic motion. This leads to the formation of compressions and rarefactions which through the medium and reach the tympanum of the ear. The tympanum membrane vibrates which leads to the sensation of hearing.

Question 2. Explain how sound is produced by your school bell.
Answer: When the bell is hit by a hammer, it begins to vibrate. The vibrations disturb the air particles which transfer the sound energy through he medium by periodic vibrations

Question 3. Why are sound waves called mechanical waves?
Answer: Since sound waves require a material medium for their propagation, they are called as mechanical waves

Question 4. Suppose you ad your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer: Since the moon is a barren place ie no atmosphere or a material medium, sound can’t travel in the moon. Hence friend cannot hear the sound

Question 5. What are the wavelength, frequency, time period ad amplitude of the sound wave?
Answer:
Wavelength(λ): For a transverse wave, the wavelength is the distance between two consecutive crests or troughs. For a longitudinal wave, it is the distance between two consecutive compressions or troughs. In general, it is the distance travelled by a wave in one period

Frequency(v): The number of vibrations executed by a particle of a medium or the number of waves produced by a source. It is measured in hertz (Hz)

Time — period (T): Time is taken by a particle to complete one vibration

Amplitude: Maximum displacement of a particle in vibration

Question 6. How are the wavelength and frequency of a sound wave related to its speed?
Answer: v = v λ where v = speed, v = frequency and λ = wavelength of sound waves

Question 7. Calculate the wavelength of a sound wave whose frequency is 220Hz and speed is 440m/s in a given medium.
Answer: λ= v/v = 440/220 2m

Question 8. A person is listening to a tone of 500Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Answer: Time interval between two successive compressions = 1/ f= 1/500 = m0.002sec.

Question 9. Which wave property determines
1) loudness
2) pitch
Answer: Intensity determines loudness and frequency determines the pitch

Question 10. Guess which sound has a higher pitch: guitar or car horn?
Answer: Guitar sound has a higher pitch than a car horn

Question 11. Distinguish between loudness and intensity of sound.
Answer:
Intensity:
1. It is the amount of sound energy incident per unit area per second
2. It is independent of the condition of the human ear

Loudness:
1. It is a physiological response of ear to the sound.
2. It depends on the intensity and condition of a human ear

Question 12. In which of the three media: air, water or iron does sound travel the fastest at a particular temperature?
Answer: Iron

Question 13. An echo returned in 3s. What is the distance sensation fo hearing? The sound produced by of the reflecting surface from the source, vibrations of objects given that the speed of sound is 342m/s?
Answer:
Speed of the sound = 342m/s
Total time take by the sound from the source to air near a source of a listener and the listener to source =3s
Total distance travelled = speed x time
= 342x 3 = 1026m
Distance of the reflecting surface from the source
= 1/2 x Half the total distance travelled
= 1026/2 = 513 m

Class 9 Science Chapter 12 KSEEB Textbook Solutions 

Question 14. Why are the ceilings of concert halls of high curved?
Answer: It is done so to make sound after multiple object reflections reach all parts of the hall.

Question 15. What is the audible range of frequency of the average human ear?
Answer: 20Hz to 20kHz

Question 16. What is the range of frequencies associated with
1)Infrasonic sound and
2)ultrasonic sound?
Answer: For Infrasonic sound < 20Hz and for ultrasonic sound >20kHz

Question 17. A submarine emits a SONAR pulse, which returns from an obstacle underwater cliffin 1.02s. If the speed of sound in sla water is 1531m m/s, how far away is he cliff?
Answer: Total distance travelled by sound = velocity x time
2x ==. 153 x 1.02 m
x = 1531 x 1.02/2 = 780.81m
Thus distance of the cliffis 789.8m

Question 18. What is sound and how is it produced?
Answer: A sound is a form of energy which produces the sensation fo hearing. The sound produced by vibrations of objects

Question 19. Describe with the help of a diagram, how compressions rarefactions are produced in air near a source of the sound.
Answer:

Source of Sound Chapter 12

When a tuning fork vibrates, It pushes and compresses the air in front of it creating a region of high density and pressure called compression. This compression moves away from the vibrating object. When the vibrating prongs move backwards, it creates a region of lower density and low pressure called rarefaction as shown in the figure. As the prongs move back and forth rapidly, a series of compressions and rarefactions is created in the air. Thus the sound wave moves through the medium in the form of a series of compressions and rarefactions.

Question 20. Cite an experiment to show that sound needs a material medium for its propagation.
Answer: An electric bell is suspended inside an air-tight glass bell jar as shown in the figure. The bell jar is connected to a vacuum pump. When a switch is pressed, we will be able to hear the sound of the bell. Now the air from the jar is pumped out gradually. The sound of the bell becomes fainter. When the air is completely removed, we hear no sound from the jar. The experiment clearly indicates that sound requires a material medium for its propagation.

Question 21. Why is a sound wave called a longitudinal wave?
Answer: When the sound wave travels through a medium, the particles vibrate to and fro along the direction of the wave motion. Hence the sound is called a longitudinal wave

Question 22. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer: Quality or timbre

Question 23. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer: It is because the speed of sound is much less than that of light

KSEEB Solutions For Sound Short Notes 

Question 24. A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344m/s
Answer: Wavelength corresponding to 20 Hz = 344/20 = 17.2 m Wavelength corresponding to 20kHz = 344/ 20,000 = 0.0172m

Question 25. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child. Speed of sound in air = 346m/s and that in aluminium = 6420m/s)
Answer:
Time is taken by the sound in air=distance /velocity
= x /346 second
Time taken by the sound in Afuminium=x/6420
Ratio of times = (x/346) / (x/6420) = 6420/346 = 18.55

Question 26. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Frequency = number of vibrations in 1 sec = 100
Hence the number of vibrations in 1 minute
ie 60 sec = 60 x 100 = 6000

Question 27. Does sound follow the same laws of reflection as light does? Explain.
Answer: Yes. Like light, the sound gets reflected from the surface of a solid or liquid and follows the same law of reflection. As in the case of light, the direction in which the is incident and reflected make equal angles with the normal to the reflecting surface and three lie in the same plane

Question 28. When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear an echo sound on a hotter day?
Answer: No. This is because, during hotter days, the speed of sound decreases and the reflected sound returns to the listener in less than 0.1 s (The minimum time required to hear the reflection of sound)

Question 29. Give two practical applications of the reflection of sound waves.
Answer:
(1) Stethoscope is used to hear the sounds produced in the lungs and heart
(2) Megaphones which is used to send sound in a particular direction

Question 30. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g= 10 m/s2 and speed of sound = 340 m Is?
Answer:
Let s = 500m, g = 10m/s2, u = 0 and t = time taken by the stone to reach the base of the tower
Now we have s = ut + Vz gt2 => 500 = Vz (10) t2 => t2 = 100
Thus t = V100 = 10 sec
Time is taken by the sound to reach the top of the tower from the base = 500.340 = 1.47s
Thus splash of the sound is heard after 10 + 1.47
= 11. 47 sec

Question 31. A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer:
Frequency = speed/ wavelength = 33 9/.015 = 22,600Hz (Ultrasonic sound) Hence the sound is not audible because the range of frequency of audible sound is 20Hz to 20,000Hz.

Question 32. What is reverberation? How can it be reduced?
Answer:
The persistence of sound in a large hall due to multiple reflections of sound is called reverberation. It can be reduced by
(1)By covering the roof and the wall of the auditorium by sound absorbing materials
(2)Using heavy curtains

Question 33. What is the loudness of sound? What factors does it depend on?
Answer: It is a physiological response of the ear to the intensity of the sound. It depends on
(1) Intensity of the sound
(2) Sensitivity of the human ear

Question 34. Explain how bats use ultrasound to catch prey.
Answer: During its flight, a bat emits ultrasonic waves. After receiving the reflected waves, the bat understands the nature of the object and its size before preying on it.

KSEEB Class 9 Science Chapter 12 Important Questions 

Question 35. How is ultrasound used for cleaning?
Answer: The objects to be cleaned are placed in a
cleaning solution and ultrasonic waves are sent into the solution. Due to high frequency, the particles of dust, grease and dirt get detached and drop out. Thus the objects are cleaned.

Question 36. Explain the working and application of a sonar.
Answer: Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. Sonar consists of a transmitter and a detector and is installed in a boat or a ship.

Sound Chapter 12

The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector. The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound. Let the time interval between transmission and reception of ultrasound signal be t and the speed of sound through seawater be v. The total distance, 2d travelled by the ultrasound is then, 2d = v x t.

Question 37. A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Answer:
Given t = 5s and total distance travelled (2d)
3625 + 3625 = 7250m
Velocity of the sound in water = 2d /1 = 7250/5
= 1450m/s

Question 38. Explain how defects in a metal block can be detected using ultrasound.
Answer: Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect.Ultra Sound Chapter 12

Question 39. Explain how the human ear works.
Answer: The outer ear is called ‘pinna’. It collects the sound from the surroundings. The collected sound passes through the auditory canal. At the end of the auditory canal, there is a thin membrane called the ear drum or tympanic membrane. When compression of the medium reaches the eardrum the pressure on the outside of the membrane increases and forces the eardrum inward. Similarly, the eardrum moves outward when a rarefaction reaches it. In this way the eardrum vibrates.

The vibrations are amplified several times by three bones (the hammer, anvil and stirrup) in the middle ear. The middle ear transmits the amplified pressure variations received from the sound wave to the inner ear. In the inner ear, the pressure variations are turned into electrical signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.

Sound Additional Questions

One Mark Questions

Question 1. What is the principle of a stethoscope?
Answer: Multiple reflections of sound

Question 2. Name two animals that can produce ultra¬sound.
Answer: Dolphins and bats

Question 3. What is the abbreviation of SONAR?
Answer: Sound navigation and ranging

Question 4. What is the nature of sound waves in the air?
Answer: Longitudinal nature.

Question 5. Can a motion be periodic but not oscillatory?
Answer: Yes. Uniform circular motion of an object

Question 6. What does that move in wave motion?
Answer: Energy

Two Marks Questions

Question 7. What is Sonic boom?
Answer: A very sharp and loud sound produced by the shock waves produced by a source of sound moving with a speed greater than sound is called a sonic boom.

KSEEB Solutions Chapter 12 Properties Of Sound Class 9 

Question 8. What is a periodic motion? Give an example
Answer: The motion of an object which is repeated itself regularly in a fixed interval of time is called periodic motion.
Example: Motion of earth around the sun

Question 9. What is an oscillatory motion? Give an example
Answer: The to and fro motion of an object about a fixed position(mean position) is called an oscillatory motion. Example: Motion of bob of a pendulum

Question 10.Differentiate between mechanical waves and non-mechanical waves
Answer:
Mechanical waves:
1. Require a material medium for their propagation
2. Maybe longitudinal or transverse
3. Ex: Sound waves, Seismic waves
Non-Mechanical waves:
1. Do not require any material medium for propagation only transverse
Ex: Light waves, radio waves

Question 11.Mention the types of waves produced in the following cases
(1) When a stone is dropped into still pond water
(2) The weight attached to a spring is slightly pulled and released
(3) The string of Veena is plucked
Answer:
(1) Transverse waves
(2) longiudinal waves
(3) Transverse waves

Question 12. What is an echo? What should be the mini¬mum distance between the source of sound and the obstacle?
Answer: The phenomenon of repetition of sound due to reflection from a large rigid surface is called echo. The minimum distance between the source of sound and the obstacle must be 17.2m.

Question 13. Distinguish between echo and reverbera¬tion.
Answer: The repetition of sound due to reflection from an obstacle is called echo. The original sound and echo can be heard separately. Reverberation is the persistence of sound due to multiple re¬flections from many obstacles.

Question 14. What is ultrasonic or ultrasound? What are their special characteristics?
Answer: The sound whose frequencies are > 20,000 Hz are called ultrasound. They can travel along well-defined paths even through obstacles.

Question 15. How is ultrasound used to clean electronic equipment?
Answer: Objects are kept in a cleaning solution and ul¬trasonic waves are sent into the solution due to high frequency, the particles of dust, grease and dirt get detached and drop out.

Three Marks Questions

Question 16. What is echocardiography (ECG)?
Answer: The technique in which ultrasonic waves are reflected from various parts of the heart and an image of the heart is obtained on a screen is called echocardiography.

Question 17.Differentiate between Longitudinal and transverse waves
Answer:
Longitudinal waves:
1. The particles in the wave vibrate along the direction of the propagation of the wave motion.
2. The waves propagate in the form of alternate crests and troughs
3. Ex: Lightwaves
Transverse waves:
1. The particles vibrate in the direction perpendicular to the direction of the Wave motion.
2. The waves propagate in the form of compressions and rarefactions.
3. Ex: Sound waves

Question 18.Differentiate between Sound waves and light waves
Answer:
Soundwaves
1. Longitudinal in nature
2. Requires a material medium for propagation
Light waves
1. Longitudinal in nature
2. Does not require a material medium for propagation.

Sound KSEEB Class 9 Detailed Solutions 

Question 19. Give a graphical representation of a transverse wave and longitudinal wave
Answer:
Transverse waves:

Sound Transverse waves

Longitudinal Waves:

Sound Longitudinal Waves

Question 20. Ocean waves of time interval period 0.01s have a speed of 15m/s. What is the distance between a wave crest and the adjoining wave trough?
Answer:
Distance between a crest and the adjoining trough = 1/2 x  wavelength
= 1/2  x  velocity / frequency = 1/2 x 15/0.01
= 0. 075 m or 7.5cm

Question 21.Mention important medical applications of ultrasound sound
Answer: Electrocardiograph (ECG), Ultra scanning and breaking of small stones in the kidneys

Question 22. Draw a neat diagram to show the auditory parts of the human ear. Mention the functions of small bones present in the middle ear
Answer:

Sound Ear Chapter 12

The small bones present in the middle ear amplify the vibrations several times

Four Marks Questions

Question 23. Explain briefly the propagation of sound in air.
Answer: When a source produces sound, it compresses air in front of it creating a region of high pressure. This region is called compression.

Propagation of sound

This compression starts to move away from the vibrating source. This leads a region of low pres¬sure called rarefaction as shown in fig. Due to elasticity and inertia, the compressions and rarefactions are transformed to neighbouring air particles. Thus a series of compressions and air particles help sound energy to propagate through air.

Question 24. Describe Bell jar experiment to show sound requires a material medium for its propagation (OR) Describe the Bell jar experiment to show that sound cannot travel in a vacuum.
Answer:

Sound vacuum

An electric bell is suspended inside an airtight glass bell jar as shown in the figure. The bell jar is connected to a vacuum pump. The electric bell is switched and the sound is heard clearly. Start the vacuum pump and remove the air in the jar gradually, the sound becomes fainter and fainter and finally stops when the air is completely removed. Even though the same current is flowing through the electric bell, we hear no sound. This experiment clearly demonstrates that sound waves cannot travel in a vacuum.

Question 25. What are called longitudinal waves? Mention their characteristics. Give an example.
Answer: The waves in which particles of the medium vibrate in the direction of wave motion are called longitudinal waves.
Characteristics:
1. They constitute alternate compression and rarefaction
2. They propagate through solids, gases and liquids.
3. They produce pressure changes in different parts of the medium
Examples:
1. sound waves
2. Waves inside the water

Question 26. What are called transverse waves? Mention their characteristics. Given examples.
Answer: The waves in which the particles of the medium vibrate in the direction perpendicular to the direction of propagation of the medium are called transverse waves.
Characteristics:
1. The wave contains alternate crests and troughs
2. They propagate through solids and on the surface of liquids
3. They involve changes in the shape of the medium
Examples:
1. light waves
2. waves in the stretched string

Question 27. What are called mechanical waves and non-mechanical waves? How are they produced? Given examples.
Answer:
Mechanical waves: The waves which require a material medium for their propagation are called Mechanical waves.
Mechanical waves are produced due to vibrations of particles of the medium
Example:
1. sound waves
2. water waves

Non – Mechanical waves: The waves which do not require any material medium or which can propagate through a vacuum are called non-mechanical waves.
Non – mechanical waves are produced due to electric and magnetic vibrations
Example:
1. light waves
2. Radio waves.

Sound KSEEB Class 9 Question Answers 

Question 28. Define the frequency and wavelength of a sound wave. Hence derive the relation V = +ve with the usual notation.
Answer:
Frequency: The number of oscillations completed by a particle of the medium in the wave motion.
Wavelength: It is the distance travelled by the wave in one period.
Relation V = +ve
The speed of sound is defined as the distance travelled by the wave in one second Speed (v) = distance/time
t / T
Here t is the distance (wavelength) travelled by the sound wave in one period (T) of the wave
Thus V = +ve

where v = frequency =  \(\frac{1}{\text { period(T) }}\)

Question 29.
1) What is a hearing aid? How does it work?
2)Ordinary sound cannot be used to detect cracks and flaws in metal blocks why?
Answer:
1) Hearing aid is a battery-operated electronic device used by people with hearing loss
working of hearing aid:
A microphone in the aid receives sound and converts it to electrical signals. These signals are amplified by an amplifier. The amplified sound is fell into a speaker. The speaker converts electrical signals to the original sound and sends to the ear.
2) Ordinary sound waves have longer wavelengths and bend around the comers of the defective parts. Hence we cannot detect the flaws.

Sound Application Questions

Question 1. Megaphones or horn has a conical opening at the end. Why
Answer: It is designed to so to send most of the sound waves in the forward direction ie towards the audience.

Question 2. Why do some animals like dogs get disturbed before an earthquake?
Answer: Dogs can sense out the low-frequency infrasound before earth square.

Question 3. What happens to the speed of sound when the temperature is increased?
Answer: The speed of the sound increases.

Question 4. Name a phenomenon exhibited by the transverse wave but not a longitudinal wave
Answer: Polarisation

Question 5. Ultrasonic waves are used to detect cracks and flaws in metal blocks. Why.
Answer: Due to their short wave, they do not bend away from cracks.

Question 6. We can hear the sound of humming bees but cannot hear the oscillation of the pendulum.
Answer: The sound of humming bees has an audible frequency ( >20 Hz)whereas the oscillation of a pendulum has a frequency of < 30 Hz.

Question 7. Sound waves are longitudinal waves. Why
Answer:
In sound waves, the particles of the medium vibrate back and forth about their mean position. Light waves consist of an oscillating electric field and magnetic field and therefore do not require any material medium for their propagation.

Question 8. Why do we call light waves electromagnetic waves?
Answer: Light waves consists of an oscillating electric field and magnetic field and therefore do not require any material medium for their propagation.

Question 9. What type of waves are produced by prongs of a tuning force?
Answer: Longitudinal waves

Question 10. What do you mean by the acceleration due to gravity is 9.8ms’2?
Answer: It means the velocity of anybody falling freely a minute. Calculate its frequency under earth’s gravity changes by 9.8 m/s.

KSEEB Solutions For Class 9 Science Chapter 14 Natural Resources

KSEEB Solutions Class 9 Science Chapter 14 Natural Resources Important Concepts

  • Resources – Air, water, soil
  • Pollution – Air, water, soil
  • Bio-geo chemical cycles, water cycle, Nitrogen cycle
  • Carbon cycle, Oxygen Cycle, Greenhouse ef-feet, and the Ozone layer.

Natural Resources: The materials present in a natural environment and useful to living organisms are called natural resources.
Examples: Air, water, soil, minerals, plants & animals.
Biosphere: It is the life-supporting zone of the earth where the atmosphere, the hydrosphere and the lithosphere interact and make life possible.
Importance of atmosphere
1)Role of atmosphere in climate control.
2)The movement of air.
3)Rain
Air pollution: It is an undesirable change in the physical, chemical or biological characteristics of air.
Effects of Air pollution: Air pollution affects the respiratory system causing breathing difficulties and diseases such as asthma, lung cancer, tuberculosis and pneumonia.

Class 9 Social ScienceClass 9 ScienceClass 9 Maths

 

Read and Learn More KSEEB Solutions for Class 9 Science 

Acid Rain: The rainwater contains a mixture of sulphuric acid and Nitric acid with a pH of less than 5.6.
Water-A wonder liquid: The oceans, rivers, streams, lakes, ponds, pools, polar ice caps, water vapour etc., collectively form the hydrosphere.
Water pollution: An undesirable change in the physical, biological or chemical qualities of water that adversely affects the aquatic life, and makes water less fit or unfit for use.
Harmful effects of water pollution: The substances like fertilizers and pesticides used in farming, and toxic metals used by industries could be poisonous. Industrial and household waste reduce the dissolved oxygen in water bodies, thereby affecting aquatic life.
Soil: It forms the upper surface of the land and supports plant growth. Soil is the layer of unconsolidated particles derived from weathered rock, organic matter, water and air.
Soil pollution: The contamination of soil (or land) with solid waste, chemicals, fertilizers and pesticides, reducing its fertility is called soil pollution.
Harmful effects of soil pollution: Excessive use of fertilizers and pesticides pollute the soil, affect its fertility, the soil thus, maybe- come acidic or alkaline.
Biogeochemical cycles: The eye he flows of nutrients between non-living environments and living organisms is known as biogeochemical cycles. It is the process of transfer and circulation of essential chemical nutrients such as carbon, hydrogen, oxygen and nitrogen in a biosphere.
Water cycle: It is the whole process in which water evaporates and falls on the land as rain and later flows back into the sea via rivers. Plants also release water to the atmosphere through transpiration. Water is also released into the atmosphere from rivers and oceans by evaporation.
Nitrogen cycle: The source of nitrogen is an atmosphere which contains 78% of the nitrogen in the form of gas. Nitrogen is an essential nutrient for all life forms. The nitrogen cycle in the biosphere involves the following important steps.
1)Nitrogen fixation
2)Ammonification
3)Nitrification
4)Denitrification
Carbon cycle: Carbon is found in various forms on the earth.
1)As carbon dioxide in the atmosphere.
2)As carbonates in various minerals.
3)As fossil fuels like coal, petroleum and natural gas. Plants utilise atmospheric carbon dioxide in photosynthesis to produce carbohydrates, which are taken by herbivores and then pass through carnivores.
Oxygen cycle: Oxygen forms about 21 per cent of the atmospheric gases. It is also present in dissolved form in water bodies and helps in the survival of aquatic life. The oxygen cycle maintains the level of oxygen in the atmosphere. Oxygen from the atmosphere is used up in three processes namely combustion, respiration and in the formation of oxides of nitrogen. Green plants are the major source of oxygen in the atmosphere.
The greenhouse effect: It is an effect occurring in the atmosphere because of the presence of greenhouse gases like carbon dioxide, methane, nitrous oxide, ozone etc., that absorb infrared radiation, thereby increasing the global temperature.
Ozone layer: This is a layer of ozone (03) surrounding the earth. It protects the earth from harmful radiation like U. V. radiation. The (CFCs) are released into the air it accumulates in the upper atmosphere and reacts with ozone resulting in the reduction of the ozone layer by forming a hole.

KSEEB Solutions For Class 9 Science Chapter 14 Natural Resources

Natural Resources Exercises

Question 1. Why is the atmosphere essential for life?
Answer: The atmosphere acts as a protective blanket for organisms to exist. It keeps the average temperature of the earth fairly steady during the day and even during the course of whole year. The atmosphere contains all the important gases which are required for sustaining life on earth.

Question 2. Why is water essential for life?
Answer: Organisms need water because it plays a vital role in the metabolic reactions taking place within an organism’s body. It acts as a universal solvent, providing a medium for chemical reactions to occur.

Natural Resources KSEEB Class 9 Question Answers 

Question 3. How are living organisms dependent on the soil? Are organisms that live in water totally independent of soil as a resource?
Answer: Soil is a complex mixture comprising of minerals, organic matter, water, air and living organisms. Soil provides a natural habitat for different organisms and also provides nutrients to the plants for their growth and development. Aquatic organisms dependent on the soil as a resource because decomposers present in the bottom sediments of water bodies decompose dead, decaying organic matter into simple, in organic substances.

Question 4. You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather?
Answer: Weather report can be recorded by information such as direction and speed of wind, temperature, humidity and patterns of cloud formation.

Question 5. We know that many human activities lead to increasing levels of pollution of the air, water-bodies and soil -• Do you think that isolating these activities to specific and limited areas would help in reducing pollution?
Answer: Human activities lead to increasing the levels of pollution of the air, water bodies and soil. Isolating such activities to specific and limited areas may help in reducing pollution of the air, water bodies and soil. Isolating sub-activities to specific and limited areas may help in reducing pollution.If we follow the safety measures we can check or stop the pollution.

Question 6. Write a note on how forests influence the quality of air, soil and water resources.
Answer: Forests influence the quality of air, soil and water resources in the following ways.
Air:
Forests help in minimising the level of carbon dioxide in the atmosphere. Plants maintain the oxygen balance in the atmosphere.
Soil:
The roots of trees prevent erosion of topsoil by holding the soil particles tightly.
Water:
Forests help in maintaining the water cycle as well as the water resources of the earth.

Natural Resources Textual Questions

Question 1. How is our atmosphere different from the atmosphere on venus and mars?
Answer: On planets Venus and mars, carbon dioxide is the major constitute but nitrogen, oxygen and water vapour are absent, so life does not exist. On our planet earth, the atmosphere contains a mixture of many gases like oxygen, nitrogen, and water vapour hence life exist.

Question 2. How does the atmosphere act as a blanket?
Answer: The atmosphere keeps the average temperature of the earth fairly steady during the day and even during the course of the whole year.

Question 3.What cause winds?
Answer: The movement of air from one region to another creates winds when the solar radiations fall on the earth, some are absorbed and a majority of these are reflected back or re-radiated by the land and water bodies. These reflected or re-radiated solar radiations heat up the atmosphere from below. As a result convection currents are set up in the air. But since land gets heated faster than the water, the air above the land gets heated faster then the air over water bodies. During the day, the air over land gets heated faster and starts rising, creating a low pressure below. As a result the air over the sea moves into this region of low pressure.

Question 4. How are clouds formed?
Answer: The water evaporates due to the Heating up of water bodies and other biological activities. The air also heats and rises, on rising, it expands and cools to form tiny droplets. These droplets grow bigger, expand and form clouds, and they fall down in the form of rain.

Question 5. List any three human activities that you think would lead to air pollution.
Answer:
The human activities that leads to air pollution are,
1)Burning of fossil fuels.
2)Smoke from automobiles.
3)Forest fires, excessive use of chloroform carbons and industrialisation.

Question 6. Why do organisms need water?
Answer:
Organisms need water for the following reasons.
1)It plays a vital role in the metabolic reactions taking place in the organisms.
2)It transports the substances from one part of the body to the other in the dissolved form.
3)It helps in the digestion of food and absorption of nutrients in the blood.

Question 7. What is the major source of fresh water in the city/town/village where you live?
Answer: The major source of fresh water in the city/village/town were we live is underground water.

Question 8. Do you know of any activity which may be polluting this water source?
Answer: Sewage and industrial wastes are the major sources of water pollution.

Class 9 Science Chapter 14 KSEEB Textbook Solutions 

Question 9. How is soil formed?
Answer: The rocks are broken down by various physical, chemical and biological processes. The breakdown of bigger rocks into small, file soil particles is called weathering. Even the sun, wind, water also helps in soil formation.

Question 10. What is soil erosion?
Answer: The removal of topsoil which is rich in humus and nutrients by flowing water or wind is called soil erosion.

Question 11. What are the methods of preventing or reducing soil erosion?
Answer:
Soil erosion can be prevented by.
1)Afforestation
2)Sowing grasses
3)Terrace forming
4)Preventing overgrazing.

Question 12. What are the different states in which water is found during the water cycle?
Answer:
Water exists in all three states of matter during the water cycle.
1)Gaseous state “Evaporation of water from the water bodies in the form of water vapour.
2)liquid state – Water vapour condenses and forms rain.
3)Solid state – Sometimes the freezing of liquid droplets in the snow.

Question 13. Name two biologically important compounds that contain both oxygen and Nitrogen.
Answer:
1)Nucleic acids (DNA and RNA)
2)Proteins

Question 14. List any three human activities which would lead to an increase in the carbon dioxide content of air.
Answer:
1)Burning of fossil fuels.
2)Industries and automobiles
3)Deforestation.

Question 15. What is the greenhouse effect?
Answer: Some greenhouse gases like co2 prevent the escape of heat from the earth. An increase in the atmosphere would cause the average temperatures to increase worldwide and this is called the greenhouse effect.

Question 16. What are the two forms of oxygen found in the atmosphere?
Answer:
1) oxygen (o2)
2) Ozone (o3)
The other sources are oxides and biological molecules.

Natural Resources Additional Questions

Question 1. What are biodegradable pollutants?
Answer: Pollutants which can be decomposed by microbial activity.
Example: Organic wastes like traits of vegetables.

Question 2. Write any two uses of carbon dioxide gas
Answer:
1)Carbon dioxide is fixed by plants to prepare food by photosynthesis.
2)Carbon dioxide can trap heat and prevent its escape from the atmosphere of the earth, maintaining temperature.

Question 3. Name two types of biogeochemical cycles.
Answer:
1)Gaseous cycle.
Example: Oxygen
2)Sedimentary cycle.
Example: Phosphorous

Question 4. What is biomagnification?
Answer: The phenomenon of an increase in the concentration of harmful non-biodegradable substances in the body of living organisms at each tropic level of the food chain is called biomagnification.

Question 5. What is eutrophication?
Answer: It is the process in which excessive growth of algal (algal bloom) occurs as a result of a high content of nutrients (nitrates and phosphates) in the water body. Eutrophication leads to the depletion of dissolved oxygen in water resulting in the killing of aquatic organisms.

Question 6. Name two acids that are usually present in acid rain.
Answer: Sulphuric acid and Nitric acid

Question 7. How are CFCS harmful for the environment and living beings?
Answer: The CFCs are not degraded by any biological process. It reacts with the ozone layer and causes hole in the ozone layer.

Question 8. What causes the movement of air?
Answer: The movement of air is caused by the uneven heating of the atmosphere in different regions of the earth.

KSEEB Solutions For Natural Resources Short Notes 

Question 9. Write the biotic and abiotic components of our environment.
Answer: Biotic components are plants, Animals and microorganisms Abiotic components are land, air, water etc.

Question 10. What is the atmosphere? List its four concentric layers.
Answer: The envelope of air that surrounds our planet earth is called the atmosphere. The four main concentric layers of the atmosphere are the troposphere, stratosphere, mesosphere and thermosphere.

Question 11. What is smog? Mention its harmful effect.
Answer: Smoke containing fog particles result in smog. It causes breathing problems.

Natural Resources High-Order Thinking Questions

Question 1. Why do not lichens occur in Bengaluru whereas they commonly grow in Darjeeling?
Answer: Lichens act as bioindicators o fair pollution and are sensitive to sulphur dioxide. Bengaluru has the maximum number of vehicles vehicular exhaust has increased concentrations of sulphur dioxide and it kills lichens.

Question 2. State reasons for the following:
1)Excess burning of coal causes the greenhouse effect.
2)Temperature ranges from – 190°C to 110°C on the surface of the moon.
Answer:
1) Excess burning of coal produces greenhouse gases like carbon dioxide methane etc. These gases have a tendency to trap the beat of the sun thereby causing the greenhouse effect,
2)The moon has no atmosphere. The atmosphere helps to regulate the temperature. In absence of an atmosphere, the temperature on the surface of the moon ranges from 190°C – 110°C

Question 3. What is nitrogen fixation? Why do plants need to fix nitrogen?
Answer: The conversion of atmospheric nitrogen into oxides of nitrogen is called nitrogen fixation, plants cannot absorb atmospheric nitrogen directly so they need to fix nitrogen. Rhizobium and blue-green algae fix nitrogen.

Question 4. There is mass mortality of fish in a pond. What may be the reasons?
Answer:
1)Thermal pollution
2)Addition of toxic compound in water.
3)Addition of pollution.

Question 5. Justify “Dust is a pollutant”?
Answer: Dust particles remain suspended in the air and can cause allergies and other respiratory diseases.

Question 6. Why do people love to fly kites near the seashore?
Answer: Due to the differential heating of land and water during the day there is a movement of air from sea to land. It helps in flying the kite high and the air provides relief near the seashore.

Question 7. Why are root nodules useful for plants?
Answer: Root nodules of leguminous plants have nitrogen-fixing bacteria Rhizobium.

Natural Resources Unit Test

Question 1. When we breathe in air, nitrogen also goes inside along with oxygen. What is the fate of this nitrogen______
1)It moves along with oxygen into the cells.
2)It comes out with the carbon dioxide during exhalation
3)It is absorbed only by the nasal cells.
4)Nitrogen concentration is already more in the cells so it is not at all absorbed.
Answer:  (2) It comes out with the carbon dioxide during exhalation

Question 2. Air is a mixture of gases with the following gas in maximum percentage_______
1)Nitrogen
2)Oxygen
3)Hydrogen
4)Carbon dioxide
Answer: (1) Nitrogen

Question 3. When water mixes with carbon dioxide in the air it forms______
1)sulphuric acid
2)carbonic acid
3)Hydrochloric acid
4)Ozone
Answer: (2) carbonic acid

Question 4. In which layer of atmospheric ozone is maximumly concentrated________
1)Troposphere
2)Stratosphere
3)Ionosphere
4)Mesosphere
Answer: (2) Stratosphere

Natural Resources Answer the following Questions

One Mark

Question 1. What keeps the temperature on earth steady?
Answer: Air prevents sudden increases in temperature during day and slows down the loss of heat at night.

Question 2. What is global warming?
Answer: Increase in atmospheric temperature due to increase in carbon dioxide.

Question 3. What is rainwater harvesting?
Answer: The method used to store rainwater by making special water harvesting structures.

KSEEB Class 9 Science Chapter 14 Important Questions 

Question 4. List the name of chemicals whose biomagnification result in the following diseases in human.
1)Minamata disease
2)Itai Itai disease
Answer:
1)Mercury
2)Cadmium

Question 5. Name any two climatic events that take place in the atmosphere
Answer:
1)Cloud formation
2)Convection currents – Winds

Question 6. Name the winds which bring rain in India.
Answer:
1)South-west monsoon
2) North-east monsoon.

Two Marks

Question 1. State any two harmful effects each of
1)Air pollution
2)Water pollution
Answer:
1)Harmful effects of air pollution are respiratory problems, Global warming and acid rain.
2)Harmful effects of water pollution are waterborne diseases such as typhoid cholera, and Eutrophication.

Question 2. State the harmful effects of ozone depletion.
Answer: Ozone prevents the harmful radiation of sun from reaching the surface of the earth. Depletion of this layer may harm many life forms by causing skin cancer, cataract in the eyes and cause global warming.

Question 3. What are the various forms in which oxygen is available?
Answer:
1) In the atmosphere, oxygen is found in its elemental form.
2) In combined form, it occurs as carbon dioxide.
3) It is also found in carbohydrates, proteins of Organic Molecule fats.

Question 4. Define humus and state its function.
Answer: The dark-coloured, partially decayed organic matter found in the top layer of soil is called humus. Humus makes the soil porous which helps the soil to increase its water-holding capacity and the content of the air.

Natural Resources Schematic Representation

1. Nitrogen cycle in nature.

Nitrogen cycle in nature Chapter 14

2. Carbon cycle in nature

 

 

 

 

3. Oxygen cycle in nature

Chapter 14 Oxygen cycle in nature

 

KSEEB Solutions For Class 9 Science Chapter 15 Improvement In Food Resources

KSEEB Solutions For Class 9 Science Chapter 15 Improvement In Food Resources Important Concepts

Crop yields, crop variety, crop production manure and fertilizers, Irrigation, cropping patterns. Animal husbandry – cattle farming, poultry farming, Fish production, Beekeeping.

Crop production: Crops are plants cultivated by human beings for food, fodder, and other materials. The important types of crops are. Cereal crops – Wheat, rice, maize, Pulses – Pea, Greengram, Oil seed – Groundnut, soybean
Crop seasons:
1)Kharif season -These crops are grown in the rainy season.
Example: Paddy, soybean.
2)Rabi season – These crops are grown in the winter season.
Example: Wheat, peas.

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Class 9 Social ScienceClass 9 ScienceClass 9 Maths

 

Macronutrients: The essential elements utilized by plants relatively in large quantities.
Examples: Nitrogen, phosphorous, potassium, and calcium.
Micronutrients: The essential elements utilized by plants in small quantities.
Examples: Manganese, Boron, Zinc, and Copper.
Manures: Manures are organic substances obtained through the decomposition of plant waste and animal excreta. Manure can be classified into
1)compost
2)Vermicompost
3)Green manure.
Fertilizers: Fertilisers are commercially produced plant nutrients.
Types of fertilizers:
1)Nitrogenous fertilizers
2)Phosphatic fertilizers
3)potassic fertilizers
Organic Farming: It is a farming system with minimal or no use of chemicals as fertilizers, herbicides, pesticides, etc., and with a maximum input of organic manures, recycled farm wastes, etc.
Irrigation: It is the process of supplying water to crop plants in the fields by means of canals, reservoirs, wells, tube wells, etc.,
Mixed cropping: It is the practice of growing two or more crops simultaneously on the same piece of land. Eg. Wheat + gram, ground nut T sunflower.
Intercropping: It involves growing two or more crops simultaneously on the same field in a definite pattern. A few rows of one crop alternate with a few rows of a second crop.
Example : soybean + maize + cowpea
Crop rotation: The growing of different crops on a piece of land in a preplanned succession is known as crop rotation.
Pest: Any destructive organism that causes great economic damage to crop plants.
Example:  Insects, mites, etc.,
Pesticide: It refers to a chemical that is used to kill a pest organism.

  1. Insecticides – killing the insects
  2. Weedicides – killing the weeds
  3. Fungicides – killing the fungi
  4. Rodenticides – killing rodents

Weeds: They are small-sized unwanted plants that grow along with a cultivated crop in a field.
Livestock: It refers to the domestic animals kept in use for milk, and flesh and includes cattle, buffaloes, sheep, and goats.
Animal husbandry: Animal husbandry is the scientific management of animal livestock. It includes various aspects such as feeding, breeding, and disease control.
Cattle farming: Cattle husbandry is done for two purposes milk and drought labor. Milk-producing females are called milch animals. Draught animals are used for farm labor.
Breeding: It means ‘to reproduce’ and is done to obtain animals with desired characteristics.
Poultry: It is the branch of animal husbandry concerned with rearing birds for eggs and meat. Poultry practices required good care for food, shelter, and disease control.
Fish production: There are two ways of obtaining fish. One is from natural resources, which is called capture fishing. The other way is by fish farming, which is called culture fishery.
Bee-keeping: Bee-keeping or Apiculture is the rearing, care, and management of honey bees for obtaining honey, wax, and other substances.

KSEEB Solutions For Class 9 Science Chapter 15 Improvement In Food Resources

Improvement In Food Resources Exercises

Question 1. Explain any one method of crop production which ensures high yield.
Answer: Intercropping is one of the methods of crop production which ensures high yield because it ensures maximum utilization of the nutrients supplied and also prevents pests and diseases from spreading to all the plants belonging to one crop infield. In this way, both crops can give better returns.

Question 2. Why are manures and fertilizers used in fields?
Answer: Manures add a great amount of organic matter in the form of humus in the soil. Fertilizers are very rich in plant nutrients such as nitrogen, phosphorous, and potassium.

Improvement In Food Resources KSEEB Class 9 Question Answers 

Question 3. What are the advantages of intercropping and crop rotation?
Answer:
Advantages of using intercropping:
1)It helps to maintain soil fertility.
2)Crops can easily be harvested and trashed separately.
Advantages of crop rotation:
1)It avoids the depletion of a particular nutrient from the soil.
2)It helps in weed control.

Question 4. What is genetic manipulation? How is it useful in agricultural practices?
Answer: Genetic manipulation is a process of transferring genes or characteristics that are desirable from one plant to another plant for the production of varieties with desirable characteristics. Genetic manipulation is useful in increasing yield, better quality, and desirable characteristics.

Question 5. How do storage grain losses occur?
Answer:
There are mainly two types of factors
1)Biotic factors — Insects, rodents, birds & microorganisms.
2)Abiotic factors Moisture content, temperature, and humidity.

Question 6. How do good animal husbandry practices benefit farmers?
Answer:
Advantages of animal husbandry practices are.
1)Increasing the yield of foodstuffs such as milk, eggs, and meat.
2)It is beneficial for the farmers as increased yield brings more income to the farmer and raises his living standard.

Question 7. What are the benefits of cattle farming?
Answer:
Cattle fanning is beneficial in the following ways
1)Good quality of meat, fiber can be obtained.
2) Milk-yielding animals and a good breed of draught animals can be obtained.

Question 8. For increasing production, what is common in poultry, fisheries, and bee-keeping?
Answer: Through cross-breeding, the production of poultry, fisheries, and beekeeping can be increased.

Question 9. How do you differentiate between capture fishing, mariculture, and aquaculture?
Answer:
Capture
fishing:
It is the process of obtaining fish from natural resources such as rivers, ponds, canals, etc.
Mariculture:
It is a practice of a culture of marine fish varieties in the open sea. 
Aquaculture:
It is the production of fish from freshwater resources.

Improvement In Food Resource Textual Questions

Question 1. What do we get from cereals, pulses, fruits, and vegetables?
Answer: Cereals give carbohydrates which provide energy. Pulses give proteins that build our bodies. Vegetables and fruits provide all essential minerals.

Question 2. How do biotic and abiotic factors affect crop production?
Answer:
Biotic and abiotic factors affect poor crop production by the following factors.
1)Weight loss
2)Poor germination ability
3)Infestation of insects.
4)Degradation in quality.

Class 9 Science Chapter 15 KSEEB Textbook Solutions 

Question 3. What are the desirable agronomic characteristics for crop improvements?
Answer:
Desirable agronomic characteristics are.
1)Tallness and profuse branching are desirable characteristics for fodder crops.
2)Dwarfness is desired in cereals so that less nutrients are consumed by these crops.

Question 4. What are macronutrients and why are they called macronutrients?
Answer: Macronutrients are essential elements utilized by plants relatively in large quantities. They require for the plants in greater amounts for proper growth and development.

Question 5. How do plants get nutrients?
Answer: The nutrients which are found in the soil get dissolved in the water and is absorbed by the roots of a plant

Question 6. Compare the use of manure and fertilizers in maintaining soil fertility.
Answer:
Manure:
1)Manure is a natural substance
2)It adds a great amount of organic matter
3)It is not nutrient specific
Fertilizers:
1)A fertilizer is a man-made substance.
2)It does not add any humus to the soil.
3)It is nutrient specific

Question 7. Which of the following conditions will give the most benefits? Why?
1)Farmers who use high-quality seeds do not adopt irrigation or use fertilizers.
2)Farmers use ordinary seeds, adopt irrigation and use fertilizers.
3)Farmers use quality seeds adopt irrigation use fertilizers and use crop protection measures.
Answer: Option (c) is the right answer.
The use of only quality seeds and irrigation and fertilizers are not sufficient for better yield, but also crop should be protected by biotic and abiotic factors.

Question 8. Why should preventive measures and biological control methods be preferred for protecting crops?
Answer: Diseases are spread by pathogens so to get rid of pathogens some preventive measures, as well as biological methods, are preferred.
Example: To kill the pathogen biological method will be used without affecting the plant or soil quality.

Question 9. What factors may be responsible for losses of grains during storage?
Answer:
1) Biotic factors – Insects, birds, and rodents
2) Abiotic factors — moisture content, temperature, and humidity.

Question 10. Which method is commonly used for improving cattle breeds and why?
Answer:
Cross-breeding of cattle is commonly used for improving cattle breeds because.
1)It helps in getting the desired traits.
2)To get disease-resistant and high-yielding cattle breeds.

Question 11. Discuss the implications of the following statement. “It is interesting to note that poultry is India’s most efficient converter of low-fiber food stuff (which is unfit for human consumption) into highly nutritious animal protein food.
Answer: Poultry farming is to raise domestic fowl for egg production and chicken meat. The poultry birds are also efficient converters of agricultural byproducts.

Question 12. What management practices are common in dairy and poultry farming?
Answer:

  • Providing proper shelter.
  • Feeding
  • Caring for animal health.

Question 13. What are the differences between broilers and layers and in their management?
Answer: The poultry bird groomed for obtaining meat is called a broiler. The egg-laying poultry bird is called a layer. The proper nutrition and environmental requirements of broilers are somewhat different from those of egg layers.

Question 14. How are fish obtained?
Answer: The fish can be obtained by
1)Capturing
2)Culturing

KSEEB Solutions For Improvement In Food Resources Short Notes 

Question 15. What are the advantages of composite fish culture?
Answer: The composite fish culture is a combination of five or six fish species is used in a single pond.
Advantages
1. They do not compete for food among themselves.
2. The (fishes) have different types of food habits
3. Fishes feed in different zones.

Question 16. What are the desirable characteristics of the varieties suitable for honey production?
Answer:
1)The variety of bees should be able to collect a large amount of honey.
2)They (Bee) should stay in a given beehive for a long period.

Question 17. What is pasturage and how is it related to honey production?
Answer: Pasturage means the flowers available to the bees for nectar and pollen collection. In addition to an adequate quantity of pasturage, the kind of flowers available will determine the taste of the honey.

Improvement In Food Resources Additional Questions

Question 1. Name the crop whose production has increased by
1)blue revolution
2) yellow revolution
Answer:
1) Blue revolution—Fish production
2)Yellow revolution—Oil production

Question 2. Is organic farming beneficial and why?
Answer: Yes organic fanning is very beneficial because it is a farming system with no use of chemicals with a maximum input of organic manures, recycled from waste.

Question 3. Why removal of weeds is necessary?
Answer: Weeds should be removed because it takes up nutrients from soil and reduce the growth of crops.

Question 4. Which is the most advantageous fish culture system?
Answer: Composite fish culture because fishes do not compete for food among them.

Question 5. How genetically modified crops can be obtained?
Answer: Genetically modified crops can be obtained by introducing a gene that would provide the desired characteristics.

Question 6. How does catla differ from mrigal?
Answer: Catla is a surface feeder, while mrigal is a bottom-feeding fish.

Improvement In Food Resources High Order Thinking Questions

Question 1. On what factors does the growth of plants and flowering are dependent?
Answer: Temperature and photoperiod (duration of sunlight)

Question 2. Neem and turmeric powders are often used in grain storage.
1)What are they called?
2)What is the purpose of using need and turmeric?
Answer:
1) Neem and turmeric powders are biopesticides,
2) Neem and turmeric keep away insects rodents, fungi, bacteria, etc from the stored foods.

Question 3. A farmer had a plot just beside the bank of a river. Each time be planted Kharif crops, crops got damaged due to floods. He consulted the agricultural scientist who gave him a special variety of seeds and also advised him to practice fish farming.
1)What was the specialty of seed grains?
2)What name can be given to this type of fish farming?
Answer:
1) The special variety of seed grains are genetically modified to protect them from the damaging effect of floods,
2)Composite fish culture.

KSEEB Class 9 Science Chapter 15 Important Questions 

Question 4. What determines the quality of honey?
Answer:
1) The pasturage, i.e, the kind of flowers available.
2) Apiary location.

Question 5. Name one indigenous and one exotic breed of poultry.
Answer:
1) Indigenous—Aseel
2) Exotic—Leghorn.

Question 6. What are the desired agronomic characteristics for fodder and cereal crops?
Answer: Tallness and profuse branching are desirable characteristics for fodder crops. Dwarfhess is desired in cereals so that less nutrients are consumed by these crops.

Question 7. Excessive use of chemicals, such as insecticides and pesticides causes a threat to ecology.” Explain with reason.
Answer:
1) Excessive use of fertilizers and pesticides when washed away courses water pollution,
2) The pesticides accumulate in the soil for a long period.
3)They may lead to eutrophication,
4)They may result in biological magnification.

Improvement In Food Resources Unit Test

Question 1. To solve the food problem of the country, which among the following is necessary
1)Increased production and storage of food grains.
2)Easy access of people to the food grains
3)Proper storage of food grains.
4)All of these
Answer: (1) Increased production and storage of food grains.

Question 2. The place where bees are reared for commercial honey production is called
1)Beehive
2)Aviary
3)Apiary
4)collection unit
Answer: (3) Apiary

Question 3. The process of crossing genetically dissimilar plants of a species is called
1)Intervarietal cross
2) Interspecific cross
3)Intergeneric cross
4) Ail of these
Answer: (1) Intervarietal cross

Improvement In Food Resources Give One Word For The Following

1. Planting soybean and maize in alternative rows in the same field is called as Intercropping
2. Parthenium is commonly known as weed
3. Growing different crops on a piece of land in pre-planned succession is known as Crop rotation

Improvement In Food Resources Answer The Following Questions

one mark

Question 1. Why do organisms need food?
Answer: Food is required for body development, growth, and good health.

Question 2. Which technique is incorporated to obtain crop varieties?
Answer: Hybridisation.

Question 3. Which species of honey bee is used for the commercial production of honey?
Answer: Apis mellifera

Question 4. What is Aquaculture?
Answer: The production and management of fish is called aquaculture.

Question 5. Define the term hybridization.
Answer: Hybridisation refers to crossing between genetically dissimilar organisms.

Question 6. Give two examples of milch animals.
Answer: The cow and buffalo are milk animals.

Question 7. What is the advantage of green manure?
Answer: Green manure helps in enriching the soil with nitrogen and phosphorus.

Two And Four Marks

Question 1. Write any two methods of weed control.
Answer:
1) Weeds can be controlled by using pesticides.
2)Weeds can be removed by using a harrow, hoe or by cutting them.
3)The biological method involves growing crops like mustard along with the main crops which do not allow weeds to grow.

Question 2.What are th e ad vantages of hybridisation ?
Answer:
1) It makes crops disease resistant and a good response to fertilizers.
2) It gives a higher yield and improves product quality.

Question 3. Write any two characteristics of a storage structure for grains.
Answer:
1)Storage structures should be cleaned dried and airtight.
2)Storage structures should have inbuilt arrangements for aeration, temperature control, protection from pests, etc.

KSEEB Solutions Chapter 15 Agricultural Techniques Class 9 

Question 4. State three ways by which pests attack the plants and name the chemical used to control pests.
Answer:
1) The pests cut the root, stem, and leaf,
2)They suck the cell sap from plants.
3)They bore into the stem, root, and fruit. Pesticides should be used to control pests.

Question 5. How can poultry farming be integrated with crop production?
Answer:
1) Poultry birds can be fed with farm wastes like degraded grains and certain parts of a plant as food.
2) Bird wastes or excreta can be used as manures to the plants.

Question 6. How is fumigant different from pesticides?
Answer: Fumigant is used before grains are stored for future use. Pesticide is used after the germination of seed and it is sprayed over the crop plant.

Question 7. List three factors on which cultivation practices and crop yield depends.
Answer:
The cultivation practices and crop yield depends upon the following factors.
1)Availability of high-quality seeds.
2)Availability of waters
3)Availability of fertile soil
4)Access to new information and technologies.

KSEEB Class 9 Science Notes for Chapter 8 Motion

KSEEB Class 9 Science Notes for Chapter 8 Motion Important Concepts

Rest
A body is said to be at rest if its position does not change with respect its surroundings

Motion
A body is said to be in motion if its position changes with respect to its surroundings.

Reference Point
It is a fixed point with respect to which a body is at rest or in motion.

Distance
It is the length of the actual path travelled by a body in a given interval of time.

Displacement
It is the shortest distance between the initial and final positions of a body in a specified direction.

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Scalar Quantities
These are the physical quantities which have only magnitude and no direction. Ex: speed, mass, temperature, energy, work, etc

Vector quantities
These are physical quantities with both magnitude and a specified direction. Ex: velocity, force, acceleration, momentum, etc

Uniform motion
The motion of a body is said to be uniform if it covers equal distances in equal intervals of time however small the time intervals.

Class 9 Social ScienceClass 9 ScienceClass 9 Maths

 

Non-uniform motion
The motion of a body is said to be non-uniform if it covers unequal distances in equal intervals of time however small the time intervals

Speed
It is the distance travelled by the body divided by the time taken to cover the distance OR it is the distance travelled by a body in unit time (per second)

\(\text { Speed }=\frac{\text { dis } \tan c e}{\text { time }}\)

S.I unit is meter per second (m/s or\( \mathrm{ms}^{-1}\))

Class 9 KSEEB Science Chapter 8 Motion Notes 

KSEEB Class 9 Science Notes for Chapter 8 Motion

Uniform speed
The speed of a body is said to be uniform if the body covers equal distances in equal intervals of time.

Non-uniform speed
The speed of a body is said be non-uniform if the body covers unequal distances in equal intervals of time.

Average speed
It is the total distance travelled by a body divided by total time taken by the body to cover the distance.

\(\text { Average speed }=\frac{\text { total } \text { dis tan ce travelled }}{\text { total time taken }}\)

Velocity

  • It is a displacement of a body divided by the time taken to cover the displacement Or
  • It is the displacement of a body per second Or
  • It is the distance travelled by the body in one second in a given direction
\(\text { Velocity }=\frac{\text { displacement }}{\text { time }}\)

S .1 unit is metre per second (m/s or \(\mathrm{ms}^{-1}\))

Uniform Velocity
The velocity of a body is said to be uniform if the body covers equal distances in equal intervals of time however small the time intervals in a given direction.

Variable Velocity
The velocity of a body is said to be variable if magnitude or direction or both of magnitude and direction of velocity change.

Average velocity
It is not displacement covered by the body divided by the total time taken.

\(\text { Velocity }=\frac{\text { Net displacement }}{\text { Total time taken }}\)

Acceleration
It is the rate of change of velocity of a body with respect to time Acceleration

\(\text { Acceleration }=\frac{\text { change velocity }}{\text { time }}\)

S.I unit is  \(\mathrm{m} / \mathrm{s}^2 \text { or } \mathrm{ms}^{-2}\)

Positive Acceleration
Acceleration of a body is said to be positive if it takes place in the direction of the velocity of the body. In this case, velocity increases with time.

Negative Acceleration
Acceleration of the body is said to be negative if it takes place opposite to the direction of the velocity of the body. In this case, velocity decreases with time. It is also known as retardation or deceleration.

Uniform Acceleration
If the change in the velocity of a body in equal intervals of time is always the same, then an acceleration of the body is said to be the uniform acceleration

Variable Acceleration
If the change in the velocity of a body in equal intervals of time is not the same, then the acceleration of the body is said to be the variable acceleration

Graphs
It is a convenient method of presenting basic information of the motion of an object

Distance – Time Graph

  • It is the graph which represents the change in the position of the object with respect to time. It may be linear or non-linear
  • It is obtained by taking time (t) along the x-axis and distance (s) along the y- axis
  • The slope of the graph gives the speed of the body in a given time interval

Displacement – Time Graph

  • The graph was obtained by taking time along the x-axis and displacement along the y-axis. It is always a straight line (linear)
  • The slope of the graph gives the velocity of the body

Velocity – Time Graph

  • It represents the change in velocity with respect to time.
  • It is obtained by taking time(t) along the x-axis and velocity (v) along the y-axis
  • The graph may be linear or non-linear
  • The slope of the graph gives the acceleration of the body and the area under the curve gives the distance travelled by the body in the given interval of time.

Equations Of Motion
The motion of a body with uniform acceleration can be described by the following three equations
(1) u = v + at
(2) s=ut +1/2 at²
(3) 2as = v2 – u2

where u – Initial velocity, v = final velocity, a ~ acceleration, s = distance and t = time

Uniform circular motion

  • When a body moves in a circular path with constant speed, then the motion of the body is said to be uniform circular motion.
  • In uniform mo ion, velocity is variable due to changes in direction continuously. The magnitude of the velocity is given by
    the circumference of the circular path.
\(\mathrm{v}=\frac{\text { circumference of circular path }}{\text { time period }}=\frac{2 \pi r}{T}\)

where r=radius of the circular path and T – time taken to cover the path once.

KSEEB Science Chapter 8 Motion Explanations For Class 9 

KSEEB Class 9 Science Notes for Chapter 8 Motion

Exercises

Question 1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
Answer: Yes. Suppose a student throws a ball to a height and catches it.
Distance travelled by the ball = h + h = 2h m
Displacement = 0.

Question 2. A farmer moves along the boundary of a square field of side 10m in the 40s. What will be the magnitude of displacement of the
farmer at the end of 2 minutes 20 seconds from his position?

Motion Chapter 8Answer: Consider the square field ABCD as shown in the figure. Given that fanner covers the boundary ABCD in 40 s. If he walks for 2 minutes and 20 seconds, he will be at point C.

Thus magnitude of the displacement of the farmer
= AC
=√(AB)² + (BC)²
=√10² +10²
= √100+100
= √200
=14.14 m

Question 3. Which of the following is true for displacement?
(1)It cannot be zero
(2)Its magnitude is greater than the distance travelled by the object
Answer: Both statements are false

Question 4. Distinguish between speed and velocity
Answer:
Speed 

1. It is the distance travelled by the body in one second irrespective of direction.
2. It is a scalar quantity
3. It is always positive

Velocity
1. It is the distance travelled by a body in one second in the given direction.
2. It is a vector quantity

Question 5.Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Answer: When a body moves along a straight path in the same direction, its total path length is equal to the magnitude of the displacement. Under this condition, average velocity is equal to average speed.

Question 6. What does the Odometer of an automobile measure?
Answer: The odometer measures the distance travelled by automobile

Question 7. What does the path of an object look like when it is in uniform motion?
Answer: The path looks like a straight line

Question 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light ie 3 x \0sm/s.
Answer: Given data, t = 5 minutes 300 second and
v = 3 x 10 m/s
Distance of the spaceship (s) = v
t = 3 x 108 x 300 = 9 x 10 10 m

Question 9. When will you say a body is in (1) uniform acceleration? (2) non-uniform acceleration?
Answer:
(1) If the change in velocity in equal intervals of time is always the same, then the body is said to be in uniform acceleration
(2) If the change in velocity in equal intervals of time is not the same, then the body is said to be moving with non-uniform acceleration

Question 10. A bus decreases its speed from 80kmh1 in 5 s. Find the acceleration of the bus.
Answer:
Acceleration = (Final velocity – Initial velocity)/(time interval)
Initial velocity = 80km/h

\(\frac{80 \times 1000}{3600}=\frac{200}{9}=22.22 \mathrm{~m} / \mathrm{s} \)

Final velocity=60km/h

\(\frac{60 \times 1000}{3600}=\frac{50}{3}=16.67 \mathrm{~m} / \mathrm{s} \)

Acceleration = \(\frac{16.67-22.22}{5}=-6.47 / 5=-1.11 \mathrm{~ms}^{-2}\)

Question 11. A train starting from a railway station and moving with uniform acceleration attains a speed of 40km 1 in 10 minutes. Find its acceleration
Answer:
Final Speed= 40 km/h= \(\frac{40 \times 1000}{3600}=\frac{100}{9}=11.11 \mathrm{~m} / \mathrm{s}\)

Initial Speed = 0 km/h
Acceleration = (Final velocity – Initial velocity)/(time velocity)

\(=\frac{11.11-0}{10 \times 60}=\frac{11.11}{600}=0.019 \mathrm{~m} / \mathrm{s}^2\)

Question 12. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer: For uniform motion, the distance-time graph is a straight line. For non-uniform motion, the distance-time graph is a curved line

Question 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer: The object is at rest

KSEEB Science Notes Chapter 8 Motion With Solved Examples 

Question 14. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer: The body is moving with uniform motion ie uniform speed.

Question 15. What is the quantity which is measured by the area occupied below the velocity-time curve?
Answer: Displacement covered by the body in the given interval of time.

Question 16. A bus starting from rest moves with a uniform acceleration of 0.1 m/s2 for 2 minutes. Find
(1) the speed acquired
(2) the distance travelled

Answer: Given data: u=0, a=0.1 m/s2 t=2 minutes – 120 second
(1) The speed acquired (v) = u+at
\(=0+0.1 \times 120=12 \mathrm{~ms}^{-1}\)
(2) The distance travelled =\( u t+1 / 2 a t^2\)
\(=0 \times 120+1 / 2 \times 0.1 \times(120)^2\)
=720m

Question 17. A train is travelling at a speed of 90km/h. Brakes are applied so as to produce a uniform acceleration of 0.5 m/s2. Find how far the train will go before it is brought to rest.
Answer: Given data:
\(\begin{aligned}
&\mathrm{u}=90 \mathrm{~km} / \mathrm{h}=\frac{90 \times 1000}{3600}=25 \mathrm{~m} / \mathrm{s} \\
&\mathrm{a}=-0.5 \mathrm{~m} / \mathrm{s}^2, \mathrm{v}=0
\end{aligned}\)
We know that \(v^2-u^2=2 a s\)
\(\mathrm{s}=\frac{v^2-u^2}{2 a}=\frac{0^2-25^2}{2(-0.5)}=625 m\)

Question 18. A trolley while going down an inclined plane, has an acceleration of 2cm s’2. What will be its velocity 3s after the start?
Answer:
Given data: u = 0, a=3cm=0.03m, t=3sec.
Velocity after 3 sec = v=u+at \( =0+0.03 \times 3\)
\( =0.09 \mathrm{~m} / \mathrm{s}\)

Question 19. A racing car has a uniform acceleration of 4 m/s2. What distance will it cover in 10s after the start?
Answer:
Given data : \(\mathrm{u}=0, \mathrm{a} 4 \mathrm{~m} / \mathrm{s}^2, \mathrm{t}=10 \mathrm{~s}\)
Distance travelled after \( \begin{aligned}
&10 \mathrm{~s}=\mathrm{s}=u \mathrm{t}+1 / 2 \mathrm{at}^2 \\
&=0 \times 10+1 / 2(4)(10)^2 \\
&=200 \mathrm{~m}
\end{aligned}\)


Question 20. A stone is thrown in a vertically upward direction with a velocity of 5m/s. If the acceleration of the stone during its motion is 10ms’2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer:
Given data: \(\mathrm{u}=5 \mathrm{~m} / \mathrm{s}, \mathrm{a}=-10 \mathrm{~m} / \mathrm{s}^2\)
At the height point v=0
Thus using the formula
\(\mathrm{s}=\frac{v^2-u^2}{2 a}=\frac{0^2-5^2}{2(-10)}=\frac{25}{25}=1.25 \mathrm{~m}\)
Thus the height attained by the stone = 1.25m
Now, v=u+at \( \begin{aligned}
\Rightarrow 0=5-10 \mathrm{t} & \Rightarrow \mathrm{t}=5 / 10 \\
&=0.5 \mathrm{~s}
\end{aligned}\)


Question 21. An athlete completes one round of a circular track of diameter 200m in the 40s. What will be the distance covered and displacement at the end of Z minutes 20s?

Answer: Time is taken to complete 1 round =40s
Total time taken = 2minutes+20s=140s
Total distance covered = (circumference of circularpath*total time taken)/(time taken to complete one round)
\(=\frac{\pi d \times 140}{40} \mathrm{~m}\)
\(=\frac{22}{7} \times 200 \times \frac{7}{2}=2200 \mathrm{~m}\)
After the 140s, the athlete will be in the exactly diametrically opposite position. (Why?)
Hence displacement = 200m

Question 22. Joseph jogs from one A to the other end B of a straight 300m road in 2 minutes 30 seconds and then turns around and jogs 100m back to point Cin another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C
Answer:
Case (1):
In jogging from A to B,
Distance covered = 300m and displacement = 300m(Why?)
Time taken = 2 minutes + 30s = 150s
Average Speed = (distance)/(time) =
\(\frac{300}{150}=2 \mathrm{~m} / \mathrm{s}\)
Average velocity = (displacement)/(time) =
\( \frac{300}{150}=2 \mathrm{~m} / \mathrm{s}\)

Case (2): In jogging from A to C
Distance travelled = 300 + 100 = 400m
Displacement = AB – BC = 300-100 = 200m
Time taken = 2 min 30s + 1 minute = 210s
Average Speed =(distance)/(time) =
\(=\frac{400}{210}=1.9 \mathrm{~m} / \mathrm{s}\)
Average velocity = (displacement)/(time)=
\(=\frac{200}{210}=0.95 \mathrm{~m} / \mathrm{s}\)


Question 23. Abdul while driving to school, computes the average speed for his trip to be 20kmh~1. On his return trip along the same route, there is less traffic and the average speed is 30km What is the average speed for Abdul’s trip?

Answer: Let the distance between Abdul’s house and school = X km
Time is taken in driving to school at a speed of
\(20 \mathrm{kmph}=\frac{\text { dis } \tan c e}{\text { time }}=\frac{x}{20} h\)

Time taken in returning at a speed 30kmph =\( \frac{x}{30} h\)

Total time is taken for the whole trip =

\(\frac{x}{20}+\frac{x}{30}=\frac{5 x}{60} h\)

Average speed = (total distance)/(total time taken)

= \(\frac{2 x}{5 x / 60}=\frac{120}{5}=24 \mathrm{kmh}^{-1}\)

Free notes for KSEEB Class 9 Science Chapter 8 Motion 

Question 24. A driver of a car travelling at 52k mb1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another car driver going at 34 knlr1 in another car applies his brakes slowly and stops in 10s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after brakes were applied?
Answer:Motion 1

In the figure, AB and CD represent the speed versus time graphs for the two cars.

Distance travelled by the 1st car
\(\begin{aligned}
& =1 / 2 \times \mathrm{AB} \times \mathrm{AO} \\
& =1 / 2 \times \frac{52 \times 1000}{3600} \times 5 \\
& =36.1 \mathrm{~m}
\end{aligned}\)

Similarly, the distance travelled by the 2nd car =
\(\begin{aligned}
& =1 / 2 \times \mathrm{CO} \mathrm{CD} \\
& =1 / 2 \times \frac{34 \times 1000}{3600} \times 10
\end{aligned}\)
= 47.2 m

It is clear that the 2nd car travelled farther than the 1st car after breaks were applied.

Question 25. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graphs
and answer the following questions :

Motion(1) Which of the three is travelling the fastest?
(2) Are all three ever at the same point on the road?
(3) How has C travelled when B passes A?
(4) How far has B travelled by the time it passes C?
Answer:
We know the slope of the distance-time graph represents the speed of the object.
(1) Steeper the graph, the greatest and B travels with the fastest speed
(2) Since there is no common point of intersection of the three lines, they are not at the same point ever on the road.
(3) From the graph, C has travelled a distance of about 9km
(4) Again from the graph, B has travelled about 6km

Question 26. A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10m/s2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Given data, u=0, h=20m, a = \(10 \mathrm{~m} / \mathrm{s}^2\), v = ?, t = ?
We
know that \( v^2-u^2=2 a h\)
\( v^2-0^2=2 \times 10 \times 20\)
\(v^2=400 \Rightarrow v=20 \mathrm{~m} / \mathrm{s}\)
Now we know that \( \mathrm{v}=\mathrm{u}+\mathrm{at} \Rightarrow\)
\(t=\frac{v-u}{a}=\frac{20-0}{10}=2 s\)

Question 27. The speed-time graph for a car is shown in figure 8.12.

Motion 2

(1) Find how far the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by car during the period.
(2) Which part of the graph represents the uniform motion of the car?
Answer:
(1)Along time – axis, 5 squares = 2s
Along speed – axis, 3 squares = 2m/s
Area of 15 squares = Distance
\(=2 \mathrm{~s} \times 2 \mathrm{~m} / \mathrm{s}=4 \mathrm{~m}\)
Area of 1 square = distance pf 4/15m
The number of squares under the graph between 4s and 6m/s = \(\begin{gathered}
57+1 / 2(6) \\
=60
\end{gathered}\)
Hence distance travelled by the car = 60 x 4/15 = 16m

(2) The graph after 6s represents uniform motion since the graph remains flat

KSEEB Science Notes for 9th Standard Chapter 8 Motion 

Question 28.State which of the following situations are possible and give an example for each of these
(1) an object with constant acceleration but with zero velocity
(2) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:
(1) When an object is thrown vertically upwards, at the highest point, velocity becomes zero but still it has acceleration which is equivalent to acceleration due to gravity.
(2) When an aeroplane moves horizontally, acceleration due to gravity acts on it perpendicular to the direction of the motion.

Question 29. An artificial satellite is moving in a circular orbit of radius 42250km. Calculate its speed it it takes 24 hours to revolve around the earth.
Answer:
Given data, Radius (R) = 42, 250km = 42250000m
Time period (T) = 24 h = \(24 \times 60 \times 60 \text { seconds }\)
Speed =

\(\frac{2 \pi R}{T}=\frac{2 \times 3.14 \times 42250000}{24 \times 60 \times 60}=3071 \mathrm{~m} / \mathrm{s}\)

KSEEB Class 9 Science Notes for Chapter 8 Motion

Additional Questions

One Mark

Question 1. What is motion?
Answer: A body is said to be in motion if its position changes with respect to a reference point

Question 2. What is the simplest motion?
Answer: Motion along a straight line is the simplest motion

Question 3. Rest and motion are relative terms. Why?
Answer: An object at rest for one person is in motion for other people

Question 4. Give an example of directly perceivable
motion
Answer: Motion of a car on the road

Question 5. Give an example of an indirectly perceivable motion
Answer: Motion of the air observed due to the movement of dust or leaves of branches of a tree.

Question 6. Identify a scalar and vector from the
following
(1) distance
(2) displacement
Answer: distance – a scalar and Displacement – a vector

Question 7. Given an example for a uniform motion question
Answer:
1) Motion of the earth around the sun
2) Motion of a stone thrown vertically upwards
3) Motion of a train approaching a railway station

Question 8. Give two examples of non-uniform motion
Answer: It is the ratio of the total distance travelled to the time interval to cover the distance

Class 9 Motion KSEEB Notes With Numerical Solutions 

Question 9. Define average speed
Answer: It is the ratio of net displacement to the time interval

Question 10. What do you mean by the term reference point?
Answer: An object or place specified to locate the position of an object is called a reference point.

Question 11. Define displacement of a body.
Answer: The shortest distance measured from the initial to the final position of an object is known as the displacement of a body.

Question 12. What is an odometer?
Answer: It is a device used in automobiles to measure the distance travelled.

Question 13. What is a point object?
Answer: An object whose size is very small compared to the distance travelled by the object is called a point object.

Question 14. What is the S.I. unit of displacement?
Answer: metre (m)

Question 15. Can displacement be greater than the distance of an object?
Answer: No. Displacement may be equal to or less than distance.

Question 16. How is the speed of an object measured?
Answer:
\(\text { speed }=\frac{\text { distance travelled }}{\text { total time taken }}\)

Question 17. A body moves from rest with uniform speed. What type of distance-tome graph do we set?
Answer: A straight line passing through the origin.

Question 18. What does the area under velocity – time graph represent?
Answer: The area enclosed by the velocity-time graph and the time axis represent the magnitude of the displacement.

Question 19. Under what condition average velocity of an object equal to its average speed?
Answer: If the object moves along a straight path

Question 20. What can a speed meter measure?
Answer: Instantaneous speed ie speed at the moment of observation.

Question 21. Define average velocity
Answer: It is the ratio of net displacement to total time taken.

Two Marks

Question 22. Define the uniform speed of a body
Answer: The speed of a body is said to be uniform, it covers equal distances in equal intervals of times

Question 23. Define uniform velocity
Answer: The velocity of the body is said to be uniform if it covers equal displacements in equal intervals of time in the given direction

Question 24. Define variable velocity.
Answer: When the magnitude or direction or both of velocity change then the velocity is called variable velocity.

Question 25.Mention S.I units of
(1) velocity
(2)acceleration
Answer: S.I unit of velocity is  \( \mathrm{ms}^{-1}\)and the S.I unit of acceleration is \(\mathrm{ms}^{-2}\)

Question 26.Define uniform acceleration
Answer: Acceleration of a body is said to be uniform if changes in the velocity are equal in equal intervals of time

Question 27. Give an example of u uniformly acceleration motion
Answer: Motion of the freely falling body

Question 28. Define acceleration. Mention its S.I units
Answer: Acceleration is defined as the rate of change of velocity. S. I units are \(\mathrm{ms}^{-2}\)

Question 29. What do you understand by retardation of a body? Give an example
Answer: The rate of decrease of velocity is known as retardation or deceleration.
Example: When a stone is thrown upwards, its velocity decreases.

Question 30. Mention any two natural phenomena that occur due to motion.
Answer: Sunrise, Sunset, seasons etc.,

Question 31. Give an example for directly and indirectly perceivable motion.
Answer: The motion of a vehicle is directly perceivable motion. The motion of air is an indirectly perceivable motion

Question 32. When an object is thrown vertically upwards, it rises to a height of 10m – and returns to its initial position. What is the total distance travelled by the object? What is the displacement of the object?
Answer: Total distance travelled -10+10 = 20m
Net displacement = 0

Question 33. Name the type of the following motion
1)A man jogging in C part
2)Motion of around the earth.
Answer:
1) Non-uniform motion
2)uniform motion

KSEEB Chapter 8 Motion Revision Notes For Class 9 

Question 34. When is the acceleration is taken as (a) +ve (b)-ve
Answer: When acceleration occurs in the direction of motion, then it is said to be positive. When acceleration occurs against the direction of motion, then it is said to be negative.

Question 35. The time interval between lightning and thunder is 1.5 seconds. Find the distance of lightning from the earth (speed of sound in air 343 m/s)
Answer: Distance of lightning = speed of lightning x time informed = 343 x 1.5
= 515.5m

Question 36.Draw a typical distance-time graph of an object moving with a uniform speed
Answer:Motion 3

Question 37.Draw a typic distance-time graph of an object moving hm – a uniform speed
Answer:

Motion Distance and time

Question 38. What does the slope of the following graphs indicate?
1)displacement-time graph
2)velocity-time graph
Answer: Instantaneous speed is the speed at the moment of observation.

Three Marks Questions

Question 39.Mention any one use of
1)displacement time graph
2)velocity-time graph
Answer:
1) The slope gives a velocity of the body at any given instruction.
2) Area Enclosed by the graph gives the magnitude of the displacement of the curve.

Question 40. When the acceleration of a body is said to be
1) Positive
2) negative?

Answer: When the velocity of a body increases with time, then its acceleration is said to be positive. When the velocity of a body decreases with time, then its acceleration is said to be negative.

Question 41. Can an object have a constant acceleration with zero velocity? Give an example
Answer: Yes. An object can have a constant acceleration with zero velocity.
For example, when a stone is thrown up vertically, at the highest point velocity becomes zero but acceleration will be accelerated due to gravity.

Question 42. Uniform circular motion is an accelerated motion. Justify this statement
Answer: In a uniform circular motion, though the magnitude of velocity remains constant, its direction changes continuously. Hence, uniform circular motion is an accelerated motion.

Question 43. Write equations of motion
Answer:
(1) v = u + at
(2) s = \(u t+1 / 2 a t^2\)
(3)\( v^2=u^2+2 a s\)
where u – initial velocity, v = final velocity, a — acceleration, s = distance, t = time

Question 44. A moving car is brought to rest in 30 seconds after applying brakes. If the retardation rate is 2 m/s2, find the initial velocity of the car and the distance travelled by car.
Answer: Given t = 30s, v = 0, a = – \(2 \mathrm{~m} / \mathrm{s}^2\)
Using the equation v = u + at, we get 0 = u – 2
x 30 => u = 60 m/s

Class 9 KSEEB Science Motion Notes For Quick Learning 

Question 45. Give an example for
1)accelerated motion in the direction of motion
2)accelerated motion opposite to the direction of motion
3)zero acceleration
Answer:
1) accelerated motion in the direction of motion
2)accelerated motion opposite to the direction of motion
3)zero acceleration

Question 46. The velocity-time graph for the motion of an object is given

Motion velocity and time

What type of motion the part
1)OA
2)AB
3)BC represent?
Answer:
1) OA represents uniformly accelerated motion
2)AB represents zero acceleration motion
3)non-uniformly accelerated motion.

Four Marks

Question 47.
1) Distinguish between distance and displacement.
2) Usha swims in a 90 m-long pool. She covers 180m in 1 minute by swimming from one end to another and back along the same straight path. Find the average speed and average velocity of Usha.
Answer:
Distance

  • Distance is the actual path travelled by the body
  • A scalar quantity
  • Distance is always greater or equal to dis-placement
  • The distance can be positive or zero

Displacement

  • The shortest distance between an initial and final position of the body.
  • A vector quantity
  • Displacement is always equal to less than the distance
  • Displacement can be +ve, -ve or zero.

b)\( \text { Average speed }=\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
\( =\frac{180}{60} \mathrm{~ms}^{-1}\)
\( =3 \mathrm{~ms}^{-1}\)

Question 48. Prove that the area enclosed by the veloc¬ity-time graph and the time axis is equal to the magnitude of the displacement.
Answer:

Motion Magnitude of displacement

Consider car moving with uniform velocity V. The velocity-time graph PQ is a straight line parallel to the time axis as shown in fig.

Let A and B be the position of the car at the time t. and t2 respectively. W.K.T.

distance travelled the car – velocity x time
\(=v\left(t_2-t_1\right)\)
= AC x CD
S = Area of the rectangle ABDC then the distance travelled by car is equal the area enclosed by the velocity-time graph and time axis.

Question 49. From the following velocity-time graph of a car, find the distance travelled by the car from t = 10 to t = 20

Motion time and velocit

Answer: From the above velocity-time graph, the distance travelled by the car from t = 10S to 20S is given by
S = Area of rectangle ABCD + Area of triangle ADE
\(=(A B \times B C)+1 / 2 \times A D \times D E\)
AB = 20 \(\mathrm{~ms}^{-1}\)
BC = 10s
AD = AB = 10s
ED = CE – CD
= CE – AB
= 40 – 20
= \(20 \mathrm{~ms}^{-1}\)

\(S=(20 \times 10)+1 / 2 \times 10 \times 20\)
= 200 + 100
s = 300m

Question 50.Derive the equation of motion v = u + at using the v -1 graph

Motion velocity and time graph

Answer: consider an object moving with uniform acceleration. Let ‘u’ be the initial velocity and 1 v’ be the final velocity of the object respectively.
The velocity-time graph is a straight line represented by AB. In the graph
OA = Initial velocity=u
OE – Final velocity = v
OC = Time interval = t
Draw AD || to OC
BC = BD + DC = BD + OA
substituting BC = V and OA=U
we get = BD + U
=> BD = V – U
the slope of the v -1 graph represents acceleration ie
\(a=\frac{\text { change in velocity }}{\text { time interval }}\)
\(a=\frac{v-u}{t}\)
at = v – u
\(\Rightarrow \quad v=u+\text { at }\)

 

KSEEB Science Chapter 8 Motion Explanations For Class 9 

Question 51. Derive the equation of motion s = ut + at2 using a velocity-time graph.

Motion time and velocity graph

 

Answer: consider an object moving with uniform acceleration ‘a’.
AB represents the velocity-time graph of the object in the time interval +’. In the graph
O A = Initial velocity=u
OE = Final velocity = v
OC = Time interval = t
If S is the distance travelled by the object isn’t a second, then
S = Area of trapezium OABC
=Area of the rectangle OADC + Area of the triangle ADB
\(=(\mathrm{OA} \times \mathrm{OC})+1 / 2(\mathrm{AD} \times \mathrm{BD})\)……….(1)
From the fig.
OA = u
OC = t
BC = BD + DC = BD + OA
v = BD + u \( \Rightarrow \mathrm{BD}=\mathrm{v}-\mathrm{u}\)
\(\text { acceleration }(a)=\frac{\text { change in velocity }}{\text { time }}\)
[/latex] a=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{\mathrm{BD}}{\mathrm{t}}[/latex]
[/latex] \Rightarrow \mathrm{BD}=a \mathrm{t}[/latex]
substituting these values in equation(1)
Then [/latex]\mathrm{S}=\mathrm{u} \times \mathrm{t}+1 / 2 a t^2[/latex]
\(\mathrm{S}=u t+1 / 2 a t^2\)

Question 52. Derive the equation of motion 2as = v2 – u2 using the velocity-time graph.

Motion graph

Answer: In the figure, AB represent the velocity-time graph for an object moving with uniform acceleration (a)
Let S be the distance travelled by the object in ‘t’ second.
S = Area enclosed by the v -t graph
ie S = Area of trapezium OABC
\(=1 / 2(\mathrm{OA}+\mathrm{BC}) \times \mathrm{OC}\)
Area of trapezium = \(1 / 2(a+b) h\)
\(S=1 / 2[u+v] \times t\) ………..(1)
From the velocity-time velocity ie v = u + at
\(\Rightarrow \mathrm{t}=\frac{v-u}{a}\)
substituting this value in Eqn (1)
\(\mathrm{S}=\frac{1}{2}(u+v) \times \frac{(v-u)}{a}\)
\(\Rightarrow 2 \mathrm{as}=\mathrm{u}^2-\mathrm{v}^2\)……….(2)

KSEEB Class 9 Science Notes for Chapter 8 Motion

Application Questions

Question 1. Can displacement of a body be zero?
Answer: Yes, If a body returns to its initial position, then
its displacement is said to be zero.

Question 2. The walls of your classroom are at rest. Justify.
Answer: When time passes, the walls do not change with respect to the floor or ceiling. Hence the walls at rest.

Question 3. A body moves along a circular path of radius R. What will be its distance and displacement when it sweeps 180°?
Answer: When the body sweeps 180°, it will be diametri¬cally opposite point. Hence.
\(.\text { distance }=1 / 2 \times \text { circumference }\).
\(=1 / 2 \times 2 \pi r=\pi r\)
displacement = diameter = 2r

Question 4. Can displacement be negative?
Answer: Yes, In retardation motion, displacement is said to be negative.

Free Notes For KSEEB Class 9 Science Chapter 8 Motion 

Question 5. During the rainy day, we see lightning first and then hear thunder. Why?
Answer: The speed of light is greater than the speed of sound.

Question 6. What happens to an acceleration of an object when the force acting on it is doubled?
Answer: Since acceleration and the applied force acceleration doubles when the force is doubled.

Question 7.Velocity – time graph is parallel to the x-axis. What does this mean?
Answer: This means that the object is moving with constant velocity is zero acceleration.

Question 8. Mention the type of motion of a body from the following velocity-time graph

Motion Chapter 8

Answer: 1) uniformly accelerated motion
2)non-uniform retardation
3)non-uniformly accelerated motion

Question 9. An object moves such that the magnitude of velocity does not change but the direction changes continuously. What type of motion the object has?
Answer: Uniform circular motion.

Question 10. A girl travels from A to B towards the north from B she moves to C towards the east. Draw the resultant displacement in the diagram.
Answer:

Motion chapter 9                                       Motion Chapter 10

Question 11. Name the force that keeps an object in a uniform circular motion. What happens if the force vanishes suddenly?
Answer: Centripetal force keeps the object in a uniform circular motion. If the force vanishes, then the object moves in. a direction tangent to the circular path.

 

KSEEB Class 9 SSLC Biology Chapter 1 The Fundamental Unit Of Life Notes

KSEEB Class 9 SSLC Biology Chapter 1 The Fundamental Unit Of Life Learning Objectives

  • After Completing This Chapter, You Will Be Able To
  • Recognize Cell As The Basic Structural And Functional Unit Of Living Beings;
  • Differentiate Between Prokaryotic And Eukaryotic Cells;
  • Draw And Describe The Structural Organization Of Plant And Animal Cells And State Their Functions;
  • Describe The Process By Which Food And Water Move From
  • List The Mariner In Which The Cells Of Imng Organisms
  • Exchange Substances With The External Environment
  • And Describe The Process By Which They Obtain These Substances.

IN Earlier Classes, You Have Studied That All Living Beings Arc Composed Of Cells. A Cell Is The Structural And Functional Unit Of Life. Any Function Performed By An Organism Is The Outcome Of The Activity Of The Cell. A Cell Can Exist Independently On Its Own And Perform All The Life Processes. Many Cells Come Together To Form A Tissue And Tissues Collectively Form Organs. 71ms, Every Organ In Our Body Is Made Up Of Hundreds Of Thousands Of Cells.

KSEEB Class 9 SSLC Biology Chapter 1 The Fundamental Unit Of Life Notes

Discovery Of Cell

  • Cells Were Discovered By Robert Hooke In 1665. The Invention Of Microscope Helped In The Discovery Of The Cell. Robert Hooke (1635-1703) Developed A Primitive Microscope By Using Two Lenses For Achieving Greater Magnification. In Such A Microscope, The Object To Be Seen Was Placed On A Stage Below And Light Coming From An Oil Flame Was Thrown On It By A Convex Mirror.
  • He Observed Atliin Slice Ofcork (A Substance Obtained From The Bark Of A Tree) Under His Primitive Microscope. He Observed That The Cork Slice Had A Large Number Of Compartments Joined Together In A Honcycomb- Iike Pattern. He Named These Compartments As Cells (The Word Cell Comes From A Latin Word Celia Meaning A Compartment Or A Little Room). ’Ihis Was For The First Time That Anyone Had Observed That Living Things Consist Of Separate Units Called Cells.
  • In 1674, Anton Von Leeuwenhoek (1632-1723) Made Further Improvements And Constructed A Simple Microscope That Could Magnify Up To 270 Times.

The-fundamentals-unit-of-life-Robert-hooks-microscope

Class 9 KSEEB Biology Chapter 1 notes Fundamental Unit of Life 

Cell – The Fundamental Unit Of Life

Let Us Perform The Following Activity To Find Out More About Cells.

Activity-1

  • To Prepare Temporary Stained Mount Of Onion Peel To Study Epidermal Cells
    You Will Need
  • Onion Bulb, Forceps, Watch Glass, Glass Slide, Camel Hair Brush, Covcrslip, Knife, Water, Razor Blade, Mounting Needle, Safranin Solution, Glycerine And Microscope.

Procedure

  • Take An Onion Bulb, And Discard The Brown, Outer Dry Scales. With The Help Of A Knife, Cut The Onion Bulb Into Four Pieces. Remove One Fleshy Scale.
  • Bend The Outer (Convex) Surface Of The Fleshy Scale Towards Yourself. Do Not Break It Completely So That The Two Halves Remain Attached By A Thin Transparent Strip Of Epidermal Peel.
  • Gerilly Pull Llie Broken End. You Will Find Dial Die Thin Transparent Layer Of Epidermis Is Peeled Off Easily. Using A Pair Of Forceps, Remove The Peel And Place It On A Watch Glass Containing Water. ‘Ihis Will Prevent The Peel From Getting Folded Or Dry.
  • With The Help Of A Blade, Cut A Square Piece Of This Peel And Place It In A Drop Of Water On A Clean Glass Slide With The Help Of A Fine Camel Hair Brush. Make Sure That The Peel Is Perfectly Flat On The Slide.
  • Put A Drop Of Safeanin Solution On The Ped To Stain It And Leave It For 2-3 Minutes. Put A Drop Of Glycerine Solution Over It To Prevent It From Drying With The Help Of A Mounting Needle, Cover The Peel With A Coverslip, Taking Care That There Are No Folds Or Air Bubbles In It

 

Fundamentals-unit-of-life-preparation-of-a-temporary-stained-mount-of-onion-peel

  • Observe The Slide Under The Low Magnifying Power Of The Microscope. For Better Details, Increase The Magnification And Observe Under The High Power Of The Microscope.

Observations

  • While Observing The Slide Under The Microscope, You Will Observe That The Epidermis Is One Cell Thick. These Structures Or Cells Look Similar To Each Other- Each Cell Has A Cell Wall, Cell Membrane, Nucleus, Cytoplasm And A Vacuole.

Fundamentals-unit-of-life-epidernal-cells-of-onion-peel

  • The Cells That You Observed In The Above Activity Are The Basic Building Units Of The Onion Bulb. The Cells Of Onion Peel Are Rectangular Or Polygonal In Shape .All The Cells Are Firmly Bound Together. Like Onion, All Organisms Are Made Up Of Cells. Some Organisms Arc Made Up Of One Cell While Others Arc Made Up Of Many Cells.

Microscope

  • A Microscope Is An Instrument Which Is Used To Observe Objects That Are Invisible To The Naked Eye. For Example, Plant And Animal Cells, Bacteria, Fungi And Algae Can Be Observed Under A Microscope. Tire Two Common Types Of Microscopes Arc Compound Microscope And Electron Microscope.

The Compound Microscope

  • The Ordinary Light Or Compound Or Optical Microscope Used Extensively In Laboratories These Days Is A Greatly Improved Design Of Hooke’S Microscope It Consists Of Two Lenses, Namely, The Eyepiece Lens And The Objective Lens, Which Arc Combined To Produce A Greater Magnification. The Light Microscope Has A Magnification Up To 1500 Times, Good Enough To See Cells, Larger Organelles And Bacteria.

Fundamentals-unit-of-life-light-microscope-electron-microscope

KSEEB Class 9 Biology Chapter 1 The Fundamental Unit Of Life Explained 

Electron Microscope

  • An Electron Microscope (Em) Has Much Greater Powers Of Magnification And Resolution Than Those Of An Optical (Light) Microscope. An Electron Microscope Can Resolve Points 1 Nm Apart. In This Microscope, A Beam Of Electrons Is Passed Through The Section Of A Material
    To Produce The Image. The Electron Beam Passing Through The Specimen Section Is Focused By Electromagnets And Is Projected Onto A Fluorescent Screen For Direct View Or Onto A Photographic Plate For Permanent Recording. The Resulting Photograph Is Called An Electron Micrograph.

Electron Microscope Activity-2

  • To Prepare Temporary Stained Mount Of Leaf Peels, Root Tips Of Onion And Peels Ol Onion Of Different Sizes And Observe Epidermal Cells.
    You Will Need
  • Tradescantia Leaf, Onion Bulbs Of Different Sizes, Onion Roor Tips, Forceps, Watch Glass, Glass Slide, Water, Razor Blade, Iodine Solution And Microscope

Procedure

  • For Preparing The Temporary Stained Mounts Of Peels Of Onion Bulbs Of Different Sizes, Follow The Steps Of Activity 1 As Given Before. Observe Different Sections Under A Microscope And Compare Their Structures.
  • For Preparing The Temporary Stained Mount Of Leaf Peels, Tale The Leaf Of Tradtscaniia Plant. Take Out A Small Segment Of Peel From The Lower Surface Of The Leaf With A Jerk. Make T Temporary Mount Of This Peel As Mentioned In Activity 1 And Observe Under The Microscope.
  • Similarly, You Can Prepare Temporary Stained Mount Of Onion Root Tips To Observe Epidermal Cells.

difference-between-observation-interface

Observations

  • What Do You Observe In The Above Activity? Do All Cells Look Similar With Regard To Shape And Size? Do All Cells Look Similar In Terms Of Structure? Do You Observe Any Difference Or Similarity Among Cells From Different Parts Of The Plant?

Activity-3

To Prepare A Temporary Mount Of Human Cheek Cells And Observe It Under A Microscope
You Will Need

A Clean Ice-Cream Spoon, Methylene Blue Solution, Glass Slides, Water, Needle And Microscope.

Procedure

  • Take An Icc-Crcam Spoon And Gently Scrape The Inner Lining Of Your Cheek. It Will Collect Some Viscous Transparent Material. With The Help Of A Clean Needle, Transfer This Material On A Clean Glass Slide.
  • Add A Drop Of Water To The Smear. Also Add A Drop Of Methylene Blue Solution To If. Leave The Preparation For About 1 Minute. Methylene Blue Is Used To Stain The Nucleus In A Cell Gently Place A Coverslip Over The Material With The Help Of A Needle To Avoid The Entry Of Air Bubbles. Press It Gendy In Between Folds Of A Rough Filter Paper To Remove Excess Fluid And For Uniform Distribution Of The Cells In The Mount Prepared.
  • Observe Under A Microscope And Find Out The Structural Details Of Cheek Cells.

Fundamental-unit-of-life- human-cheek-cells

Observation

What Do You observer What Is The Shape, Of The Ri*Fk You See? Draw It On The Observation Sheet.

Inference

  • Under High Power Of A Compound Microscope, You Will Observe A Darkly Stained, Oval Or Spherical Dot-Like Structure Near The Centre Of The Cell.This Is The Nucleus. The Nucleus Is Surrounded By The Cytoplasm.
  • The Cell Membrane Forms The Boundary Of A Cell Similar Structures (Nucleus) Were Also Found In The Onion Peel Cells. You Will Also Find That There Is No Large Central Vacuole Or Cell Wall As Observed In The Onion Peel (Plant) Cells.
  • On The Basis Of Above Activities We Can Summarize That:
  • A Cell Is The Structural And Functional Unit Of Living Beings.
  • It Is Capable Of Independent Existence And Performs Essential Functions Of Life.
  • A Cell Is Usually Microscopic (Ie. Invisible To The Naked Eye) In Nature.
  • It Consists Of A Mass Of Protoplasm Surrounded By A Selectively Permeable Plasma Membrane.
  • In An Animal Cell, The Plasma Membrane Is The Only Limiting Membrane. However, In Plant Cells, Fungi And Bacteria, The Cell Is Also Surrounded By A Cell Wall.
  • All Organisms Start Their Life As A Single Cell. Every Cell Has Its Own Lifespan. The Old And Worn Out Cells Are Continuously Replaced By New Cells Except Few Cells Including Neurons.
  • Many Cells Collectively Form A Tissue And Many Tissues Together Form An Organ. The Organs Work Together As A System For The Purpose Of Survival.

Class 9 SSLC Biology Chapter 1 KSEEB Key Points 

Cell Theory

  • The Cell Theory Was Proposed By German Botanist Matthias Schlcidcn (183$) And German Physician Theodore Schwann (1839). Later On In 1855, Rudolf Virchow, A German
  • Pathologists added The Phrase Ottmis Cellula-E-Cdlula Meaning All Cells Arise From Pre-Existing Cells. The Major Points Of The Cell Theory Are As Follows:
  • All Living Organisms Are Composed Of Cells.
  • A Cell Is The Basic Structural And Functional Unit Of All Living Beings.
  • All Cells Arise From Pre-Existing Cells.

Exceptions To Cell Theory

  • Viruses Arc Non-Cellular Organisms. They Do Not Have Nucleus, Cytoplasm Or Enzyme And Do Not Perform Any Life Activity. They Can Multiply Only Inside The Living Host By Taking Over Their Machinery.
  • Bacteria And Blue-Green Algae Are Not True Cells. They Do Not Have A Nuclear Membrane And Membrane Bound Cell Organelles.

Fundamental-unit-of-life-some-unicellular-organisms

Unicellular And Multicellular Organisms

  • On The Basis Of The Number Of Cells, There Are Basically
    Two Types Of Organisms – Unicellular And Multicellular.
  • Unicellular (Uni Means Single) Organisms Are Made Up Of Only One Cell. For Example, Amoeba, Chlamydomonas, Euglena, Paramecium, Trypanosoma And Bacteria Are Unicellular Organisms.
  • Multicellular (Multi Means Many) Organisms Are Made Up Of Many Cells That Group Together To Perform Many Functions Of The Body. For Example, Fungi, Plants And Animals Are Multicellular Organisms. All Multicellular Organisms Have Been Derived From A Single Cell, Zygote, Through Cell Division.

Prokaryotic And Eukaryotic Cells

On The Basis Of Their Nuclear Organization, Cells Have Been Classified Into Two Types:
• Prokaryotic Cells
• Eukaryotic Cells

Prokaryotic Cells(Gk. Pro: Before, Karyon: Nucleus)

  • Prokaryotic Cells Are Single-Celled And Lack A Nuclear Membrane. These Cells Have Primitive Organization Of Genetic Material. The Genetic Material Is Equivalent To A Single Molecule Of A Circular Dna. These Cells Have An Undefined Nuclear Region Called The Nucleoid Due To Absence Of Nuclear Membrane. These Cells Lack Several Cytoplasmic Organelles Like Mitochondria, Lysosome, Endoplasmic Reticulum, Chloroplast And Nucleolus.
    Many Of The Functions Of These Cells Are Performed By Poorly Organized Parts Of Cytoplasm. Bacteria And Blue-Green Algae Are Examples Of Prokaryotic Cells.

Eukaryotic Cells
(Gk. Eu: True, Karyotv Nucleus)

  • Eukaryotic Cells Have A Well-Defined Nuclear Membrane. In These Cells, The Genetic Material Is Made Of Two Or More Linear Dna Molecules In The Nucleus, Enclosed In A Nuclear Membrane. These Cells Thus Have A Well- Organized Nucleus. These Cells Have Well-Developed Membrane-Bound Organelles, Such As Mitochondria, Endoplasmic Reticulum, Lysosome, Chloroplast And Nucleolus. Eukaryotic Cells Occur In Plants, Animals, Fungi And Protozoa.

Cell – Shape, Size And Function

  • Different Organisms Have Cells Of Different Kind. The Shape And Size Of Cells Are Related To The Specific Functions Performed By The Organisms.

Cell Shape

  • Cells Show A Great Variation In Their Shapes (Fig. 1.8). Most Cells Have A Definite Shape. Cells May Be Spindle-Shaped – Muscle Cells, Elongated – Nerve Cells, Oval – Red Blood Corpuscles, Cuboidal – Germ Cells, Branched – Osteocytes And So On.
  • Some Cells May Not Have Any Definite Shape, For Example, Amoeba And Leucocytes (White Blood 14 Corpuscles).

Fundamental-unit-of-life- shapes-of-different-cells-in-human-body

KSEEB Biology Chapter 1 The Fundamental Unit of Life Questions And Answers 

Cell Size

  • The Smallest Known Cell Is Mycoplasma Or Pplo (Pleuropneumonia-Like Organism). Its Size Is 0.1 To 0. 5 Pm (Micrometre).
  • The Bacterial Cell Is 0.5 To 5 Mm, Human Red Blood Corpuscles Are 7 To 20 Mm, Human Liver And Kidney Cells Arc 20 To 30 Mm And Nerve Cells Are About 90 To 100 Cm In Size.

Cell Function – Division Of Labour

  • Each Cell Performs Certain Basic Functions That Are Characteristics Of All Living Beings. There Is A Division Of Labour In Multicellular Organisms And Also Within A Single Cell In Many Cases. This Means Different Parts Of The Body (Organs) Perform Different Functions. This Is Because Each Cell Has Certain Specific Components Within It Known As Cell Organelles. Each Cell Organelle Performs A Specific Function, Such As Producing Energy, Making New Materials (Proteins, Etc.), Clearing Up The Waste Material, Etc. These Organelles Together Constitute The Basic Unit, I.E. The Cell. A Cell Is Able To Perform Its Functions Because Of These Cell Organelles.

Structural Organization Of A Cell

  • You Have Just Studied That Each Cell Has Special Components Called Cell Organelles. Although Cells Of Different Organisms Differ In Structure, Cells Within The Body Of A Multicellular Organism Also Differ In Shape, Size And Functions. In Spite Of These Differences, Every Cell Shows The Same Basic Structure – Cell Membrane Or Plasma Membrane, Nucleus And Cytoplasm. The Generalized Structures Of A Plant Cell And An Animal Cell Are Given

Fundamental-unit-of-life-ultrastructure-of-a-plant-cell-and-animal-cell

Cell Membrane Or Plasma Membrane

  • Ever Cell Is Bound By A Thin Delicate Membrane Called Cell Membrane Or Plasma Membrane.
  • Plasma Membrane Is The Outermost Covering Of The Cell Which Separates The Contents Of A Cell From Its External Environment.
    We Can Sec Plasma Membrane Only With The Help Of An Electron Microscope. Structurally, Plasma Membrane Is Very Flexible. It Is Made Up Of Organic Molecules Called Lipid (Which Are Present In A Viscous Bilayer) And Protein (Within Lipid Bilayer).
  • The Flexibility Of Plasma Membrane Helps The Cell To Engulf In (Take In) Food And Other Substances From Its External Environment. This Process Of Engulfing Food And Other Material From The Outside Environment Is Known As Eudocytosis. Protozoans Like Amoeba Get Their Food Through endocytosis.

Fundamental-unit-of-life- cell-membranes

 

Functions

  • Plasma Membrane Is Selectively Permeable. Therefore, It Allows Or Permits Die Entry And Exit Of Only Selected Substances. It Prevents The Movements Of Some Other Substances Across It
  • Plasma Membrane Bounds The Semi-Fluid Content Of The Cells.
  • It Protects Die Cell From Injury And Provides An Outer Boundary To The Cell.
  • It Allows The Flow Of Materials And Information Between Different Organelles Within Die Cell As Well As Between One Cell And Another.
  • It Has Carrier Proteins For Active Transport
  • How Does Movement Of Substances Take Place In And Out Of Cells?
  • The Movement Of Substances, In And Out Of The Cells, Occurs Through The Following Processes:
  • Diffusion Osmosis

Diffusion

  • The Movement Of The Molecules Of A Substance From Die Region Oftheir Higher Concentration To The Region Of Their Lower Concentration Is Called Diffusion.
  • Importance Of Diffusion.
  • Diffusion Plays An Important Role In Gaseous Exchange Between The Cells As Well As Between The Cell And Its Outside (External) Environment.
  • It Is A Means Of The Movement Of Ions And Molecules Throughout Die Protoplast.
  • Transpiration From Stomata Occurs By Diffusion.
  • Aroma (Smell) Of Flowers Is Due To Diffusion Of Aromatic Compounds Of Flowers To Attract Pollinators.
  • Carbon Dioxide Moves Out Of Die Cell Through Diffusion: There Are Many Metabolic Activities Like Respiration Taking Place Inside The Cell. During Cellular Respiration, Waste Like Carbon Dioxide Is Produced. This C02 Accumulates In The Cell In High Concentration. On The Other Hand, Concentration Of C02 Outside The Cell Is Lower Than That Inside The Cell. As A Result, C02 Moves From A Region Of Its High Concentration (Inside Die Cell) To A Region Of Low Concentration (Outside The Cell) Through The Process Of Diffusion. This Process Continues Till An Equilibrium (Even Distribution) Is Reached. The Diffusion Slows Down As Equilibrium Is Approached.
  • Oxygen Moves Inside Die Cell Through Diffusion: Inside Die Cell, Oxygen Is Regularly Needed For Respiration And Other Metabolic Reactions. Thus, Die Concentration Of Oxygen Inside Die Cell Decreases. As A Result, Oxygen Moves From A Region Of Its High Concentration (Outside The Cell) To A Region Of Its Low Concentration (Inside The Cell).

KSEEB Biology Class 9 Chapter 1 Notes With Diagrams 

Osmosis

  • The Diffusion Of Water Molecules Through A Scmi-Pcrmeable Membrane From A Region Where Water Is More Concentrated To A Region Where It Is Less Concentrated Is Called Osmosis.
  • In Other Words, Osmosis Is The Diffusion Of Water From Its Pure State Or Dilute Solution Into A Stronger Or Concentrated Solution Through A Semi-Permeable Membrane.

Importance Of Osmosis

  • Entry Of Soil Water Into Roots Occurs Through Osmosis.
  • Cell To Cell Movement Of Water Occurs Through Osmosis.
  • Living Cells Remain Turgid By Osmosis.
  • The Stomata Open And Close In Response To Increase Or Decrease In Osmotic Pressure Of The Guard Cell.

Activity-4

  • To Study Osmosis With The Help Of An Egg Yoa Will Need
    A Hens Egg, Glass Beakers, Dilute Hydrochloric Acid, Salt
    Solution And Water

Procedure

Take A Hen’S Egg And Remove Its Shell By Placing It In Dilute Hydrochloric Acid For Some Time. Since The Egg Shell Is Mosdy Calcium Carbonate, It Dissolves In Dilute Hcl
As The Egg Shell Is Dissolved, A Thin Outer Skin
Encloses The Egg. Put The Egg In Water For 5 Minutes And Observe.
What Do You Observe?

Observation 1

You Will Observe That The Egg Swells Because Water Passes Into It By Osmosis.
‘• Now, Take Another De-Shelled Egg And Place It In A Concentrated Salt Solution.
Again Observe For 5 Minutes. What Do You Observe?
Observation 2

You Will Observe That The Egg Shrinks. This Is Because Water Passes Out Of The Egg Into The Salt Solution As The Salt Solution Is More Concentrated.

Fundamental-unit-of-life- differrences-between-osmosis

 

Fundamental-unit-of-life-Behaviour-of-a-plant-cell-placed-in-solution-of-different-concentration

 

Fundamental-unit-of-life- diffusion-osmosis
Isotonic, Hypotonic, And Hypertonic Solutions

  • If You Take A Plant Cell And Place It In Solutions Having Different Concentrations, Then You Will Find That The Cell Shrinks In Hypertonic Solution, Swells In Hypotonic Solution While It Remains Unchanged In Isotonic Solution

If The Cell Is Kept In An Isotonic Solution

  • Isotonic Solution Is One Which Has The Same Water Concentration As That In The Cell. In Such A Solution, There Is No Net Movement Of Water Across The Cell Membrane. In This Case, There Is A Movement Of Water Across The Cell Membrane In Both Directions. However, Tlie Amount Of Water Entering Die Cell Is Exaedy the Same As The Amount Of Water Leaving The Cell. As A Result, There Is No Overall Or Net Movement Of Water And Cell Retains The Same Size.

Fundamental-unit-of-life-isotonic-solution

If The Cell Is Kept In A Hypertonic Solution

  • In A Hypertonic Solution, The Concentration Of Water Molecules In The Outside Medium Is Lower Than That Inside The Cell. That Means The Solution Becomes Very Concentrated. Water Moves Through The Cell Membrane In Both Directions, But More Water Leaves The Cell Than Entering It As A Result, The Cell Shrinks.

If The Cell Is Kept In Hypotonic Solution

  • In Hypotonic Solution, The Medium Surrounding The Cell Has Higher Water Concentration Than Inside The Cell. If You Place A Plant Cell In Hypotonic Solution, Then The Cell Will Swell Due To Osmosis. This Happens Because More Water Molecules From Outside Enter The Cell Than Leaving It. As A Result, The Cell Swells Up. Unicellular Organisms Living In Fresh Water And Most Plant Cells Absorb Water Through Osmosis.
  • Roots Also Absorb Water Through Osmosis. This Happens Because The Air Spaces In Root Hair Contain Cell Sap Which Has High Concentration Of Salt And Less

Activity-5

To Study Osmosis With The Help Of Dried Raisins Or Apricots

You Will Need

Dried Raisins Or Apricots, Concentrated Sugar Or Salt
Solution, Wrater And Petri Dish

Procedure

Take Few Dried Raisins Or Apricots And Place Them In Plain Water.
leave Them For A Few Hours.

Observation 1

You Will Observe That Each Raisin Or Apricot Absorbs Water And Swells When Placed In Pure Water.

Now, Place These Raisins Or Apricots In Concentrated Sugar Or Salt Solution For Some Rime.
What Do You Observe?

Observation 2

You Twill Observe That The Raisins Or Apricots Lose Water And Then Shrink.

Concentration Of Water Than Outside. As A Result, Water Enters The Root Cell Through Osmosis.

Free Notes For KSEEB Class 9 Chapter 1 Fundamental Unit Of Life 

Cell Wall

  • In Addition To Plasma Membrane, Plant Cells Arc Surrounded By A Cell Wall Also. Cell Wall Is An Outer, Rigid, Protective And Supportive Covering Of Plant Cells. ‘Ihe Cell Wall Lies Outside The Plasma Membrane. Its Thickness Varies In Different Types Of Cells. The Cell Wall In Plant Cell Is Mainly Composed Of Cellulose. Cellulose Is A Complex Substance And Provides Mechanical And Structural Strength To Plants. Cell Wall Is Also Present In Fungi And Bacteria But Not In Animal Cell

Functions

  • It Provides A Definite Shape To The Cell
  • It Protects Plasma Membrane And Internal Structures From The Attack Of Pathogens And Mechanical Injury.
  • It Counteracts The Osmotic Pressure.
  • It Provides Rigidity To The Cell.
  • A Living Plant Cell Can Lose Water Through Osmosis And As A Result There Is A Shrinkage Or Contraction Of Die Cell Contents Away From The Cell Wall. This Phenomenon Is Known As Plasmolysis. Let Us Perform The Following Activity To Learn More About Plasmolysis.

Activity-6

  • To Show With The Help Of Rlioeo Leaf That Only Living Cells Undergo Plasmolysis
    You Will Need
  • Rhoeo Leaf, Glass Slides, Concentrated Sugar Or Salt Solution, Wrater, Petri Dish And Microscope
  • Procedure
  • Take A Rhoeo Leaf And Break It To Take Out A Peel.
    Mount This Peel Of Rhoeo Leaf In Water And Examine Under High Power Of Microscope. What Do You Observe?Observation 1
  • You Will Observe Ihai The Leal’ Peel Contains Green Coloured Granules Called Chloroplnsts. The Chloroplasts Contain Chlorophyll, Which Helps In Photosynthesis.
    Now, Put A Drop Of Strong Or Concentrated Solution Of Sugar Or Salt On The Leaf Peel. Wait For A Minute And Observe. What Do You Sec?
  • Observation 2
  • You Will Observe That Plasmolysis Has Occurred And The Green Leaf Peel Cells Lose Water And Shrink Tire Cell Contents Away From The Cell Wall. This Is Because Water Leaves The Cells Through Osmosis.
    Now, Place The Leaf Peel In Boiling Water For A Few Minutes. This Will Kill The Cells.
    Mount This Peel Of Boiled Rhoeo Leaf In Water And Examine Under High Power Of Microscope.
    Put A Drop Of Strong Or Concentrated Solution Of Sugar Or Salt Solution On The Leaf Peel. Wait For A Minute And Observe. What Do You See? Did Plasmolysis Occur This Time?

Observation 3

  • You Will Observe That The Boiled Leaf Peel Did Not Show Plasmolysis. As The Plasma Membrane Is Dead, The Water Molecules Move Freely In And Out Of The Cell. This Shows That Only Living Cells Show Plasmolysis And Not The Dead Cells.

Plasmolysis

  • Shrinkage Of The Protoplast (Cell Content) Of A Cell From Its Cell Wall Under The Influence Of A Hypertonic Solution Is Called Plasmolysis. If We Place A Living Turgid Cell In A Hypertonic Solution, Withdrawal Of Water (Exosmosis) Occurs From The Central Vacuole Of The Cell As A Result, Die Size Of The Protoplasm Becomes Reduced And The Plasma Membrane Is Withdrawn From The Cell Wall.
  • Cells Of Plants Absorb So Much Water But Do Not Burst. Why?
  • Did You Know That The Cells Of Plants, Fungi And Bacteria Can Withstand Very Dilute (Hypotonic) External Media Widiout Bursting? Hus Is Because Of Cell Wall. In Hypotonic Media, Cells Take Up Water Due To Osmosis. As A Result, The Cell Swells Building Up Pressure Against The Cell Wall. At The Same Time, Cell Wall Also Exerts Equal
    Pressure Against The 3wou.Cn Cell Contents. Because Of
    The Pressure Exerted By The Cell Wall, The Cells Of Plants, Fungi And Bacreria Can Withstand Much Greater Changes In The Surrounding Medium Than The Animal Cells.

Fundamental-unit-of-life-Diagrametic-reprasentation-of-plasmolysis-in-cell

Nucleus

  • A Dense, Generally Round (Spherical) But Sometimes Cylindrical Nucleus Is Present Almost At The Centre Of A Cell. The Nucleus Contains:
  • Nuclear Envelope Or Nuclear Membrane,
  • Nucleoplasm Or Nuclear Sap,
  • Chromosomes (Chromatin Network), And Nucleolus
  • Nuclear Envelope: It Is A Double-Membraned Structure, And Contains A Large Number Of Pores. It Separates The Nucleus From The Cytoplasm. The Nuclear Pores Control The
  • Transfer Of Substances From Inside The Nucleus To Outside, I.E. To Cytoplasm.
  • Nucleoplasm: Inside Die Nuclear Envelope Is Die Nucleoplasm. It Is Transparent, Semi-Fluid And Contains A Large Number Of Enzymes Which Are Required For The Syndvesis And Functioning Of Dna, Rna, Etc.
  • Chromatin Network: In The Nucleus, Chromatin Network And Nucleolus Are Present. Chromatin Network (Gk. Chroma-Colour) Is A Tangled Fibrous Mass Of Thread-Like Structures.
  • The Chromatin Threads
  • Organize To Form Chromosomes Which Are Visible Only At The Time When The Cell Divides. Each Chromosome Consists Of Two Similar Threads Or Rod-Shaped Structures.
  • Chromosomes Are Composed Of Dna (Deoxyribonucleic Acid) Molecules And Proteins. The Information For Inheritance Of Characteristics From Parents To The Next Generation
  • Passes In The Form Of Dna. Dna Also Contains The Information For Cell Construction And Organization. Functional Segments Of Dna Are Called Genes. Genes Are The Carriers Of Heredity.
  • Nucleolus: It Is A Dense, Round Structure Attached To A Chromatin Fibre At A Specific Region. The Nucleolus May Be One Or More In Number And Is Not Bound By Any Membrane. It Is Rich In Protein And Rna (Ribonucleic Acid) Molecules.

Fundamental-unit-of-life- Nucleus

KSEEB SSLC Biology Fundamental Unit of Life Chapter 1 Revision Notes 

Functions

  • The Nucleus Controls Cell Metabolism And Other Activities Of The Cell, Hence, It Is Also Called Master Or Director Of The Cell.
  • Chromatin Part Of The Nucleus Possesses All The Genetic Information That Is Required For The Growth And Development Of The Organism, Along With Its Reproduction, Metabolism And Behaviour.
  • : Nucleus Plays A Central Role In Cellular Reproduction (Division Of Single Cell To Form Two Cells).
  • Along With Controlling The Cell Environment, Nucleus Also Directs The Chemical Activities Of The Cell. This Determines The Development And Future Form Of The Cell.

Cytoplasm

  • The Space Between The Plasma Membrane And The 20 Nucleus Is Filled By A Homogeneous, Translucent,
    Colloidal Liquid Called Cytoplasm. It Consists Of Various Inorganic And Organic Molecules, Such As Water, Salts, Proteins And A Variety Of Enzymes.
    Cytoplasm Also Contains Various Cell Organelles. These Cell Organelles Are Enclosed By Membranes. We Have Already Discussed That The Membrane-Bound Cell Organelles Are Present In Eukaryotes While They Are Absent In Prokaryotes

Functions

  • The Cytoplasm Helps In The Intracellular (Within The Cell) Distribution Of Molecules, Enzymes And Nutrients.
  • It Helps In The Exchange Of Materials Between Different Cell Organelles.
  • Biosynthesis Of Nucleotides, Proteins And Fart)’ Acids Takes Place In The Cytoplasm.
    Breaking Down Of Glucose Takes Place In The Cytoplasm.

Protoplasm

  • Protoplasm Is The Living Component Of The Cell Containing Cytoplasm And The Nucleus In A Living Cell The Chemical Composition Of Protoplasm Varies From One Cell To Another.
  • The Common Elements Found In Protoplasm Are Carbon, Hydrogen, Oxygen, Nitrogen, Iron, Phosphorus, Sulphur, Etc. Which Constitute Carbohydrates, Proteins, Fats, Minerals And Water. All The Living Components Of A Cell Lie In The Protoplasm And Perform Their Functions.

Cell Organelles

Each Cell Is Surrounded By A Cell Membrane. Cell Membrane Keeps The Contents Of The Cell Separate Front The External Environment. In Addition, Many Chemical Activities Take Place Continuously In Each Cell These Chemical Activities Are Required To Support The Complicated Structure And Functions Of Multicellular Organisms. Thus, To Perforin These Activities, The Cells Have Many Organelles That Float In The Cytoplasm. Some Organelles Are Membranous And Some Are Non- Membranous. Membranous Organelle Is A Characteristic
Feature Of Eukaryotes Only. Some Of These Organelles Are So Small That They Can Be Seen Only Under An Electron Microscope. Cell Organelles Carry Out Some Very Crucial Functions In Cells. The Different Cell Organelles That Are Found In The Cytoplasm Of The Cell Are Given

Endoplasmic Reticulum (Er)

  • Endoplasmic Reticulum (Also Known As Er) Is An Interconnected Network Of Membrane-Lined Tubes And Sheets That Run Through The Cytoplasm. Ar Some Places It Is Connected With The Plasmalemma (Plasma Membrane) As Well As With Tire Nuclear Envelope. It Looks Like Round Or Oblong Bags Or Long Tubules. Endoplasmic Reticulum Is Similar In Structure To The Plasma Membrane. The Endoplasmic Reticulum Always Forms A Network System.
  • Endoplasmic Reticulum Is Mainly Of Two Types – Smooth Er And Rough Er.
  • Rough Endoplasmic Reticulum (Rer) Has A Rough Membrane Because A Number Of Ribosomes Are Found Attached To Its Outer

Fundamental-unit-of-life- rough-endoplasmic-reticulum

Fundamental-unit-of-life- cell-organelles

Surface. Ribosomes Are Present In All Active Cells. Ribosomes Are The Site For Protein Synthesis. Endoplasmic Reticulum Sends These Proteins To Different Sites In The Cell As Per Requirement. Rough Endoplasmic Reticulum Is Very Well-Developed In Plasma Cells, Fibroblasts, Goblet Cells, Etc.

Smooth Endoplasmic Reticulum (Ser) Does Not Have Any Ribosomes. This Type Of Endoplasmic Reticulum Is Found In Liver Cells, Interstitial Cells, Adipose Cells, Muscle Cells, Etc. It Helps In The Manufacture Of Fat Molecules And Lipids. Lipids Together With Some Proteins Help In The Building Up Of Cell Membrane Through A Process Known As Membrane Biogenesis. Many Other Proteins And Lipids Function As Enzymes And Hormones.

Class 9 KSEEB Biology Chapter 1 Cell Structure And Function Notes 

Functions

  • Endoplasmic Reticulum Divides The Cytoplasm Into Small Compartments. It Gives Rigidity To The Cell, Thus, Acts As A Skeletal Framework Of The Cell.
  • It Serves As A Channel And Helps In Transporting Different Materials Such As Proteins Between Various Regions Of The Cytoplasm As Well As Between The Cytoplasm And The Nucleus. As It Helps In Intracellular Transport, It Is Also Known As The Circulatory System Of The Cell.
  • It Acts As A Cytoplasmic Framework And Provides A Large Surface Area Inside The Cells For Various Biochemical Activities.
    The Membranes Of Endoplasmic Reticulum Contain A Number Of Enzymes For Various Metabolic Activities.
  • In The Liver Cells Of Vertebrates, Smooth Endoplasmic Reticulum Helps In The Detoxification Of Many Poisons And Drugs.

Golgi Apparatus

Golgi Apparatus Was First Described By Camillo Golgi. It Is Present In All Eukaryotic Cells (Except Rbcs). Golgi Apparatus Consists Of A System Of Membrane- Bound Vesicles Which Are Arranged Somewhat Parallel To Each Other In Stacks Called Cisterns. These Membrane- Bound Vesicles Often Connect To The Membranes Of Endoplasmic Reticulum Forming Another Portion Of A 22 Complex Cellular Membrane System.

Fundamental-unit-of-life- goigi-apparatus

 

  • There Arc Three Distinct Components Visible In The Golgi Apparatus, Flattened Sacs Or Cistemae, Clusters Of Tubules And Vesicles, And Large Vesicles Or Vacuoles.
  • In Plant Cells, Golgi Apparatus Also Consists Of Many Freely Scattered Sub Units Called Dictyosomes.

Functions

  • Golgi Apparatus Helps In The Secretion Of Mucus, Enzymes And Hormones. The Material Synthesized Near The Endoplasmic Reticulum Is Transported To Various Targets Inside
  • And Outside The Cell Duough The Golgi Apparatus.
  • It Helps In The Storage, Modification And Packaging Of Secretory Products In Die Vesicles.
  • In Some Cases, Golgi Apparatus Also Helps In The Manufacture Of Complex Sugars From Simple Sugars. The Golgi Apparatus Also Helps In Die Formation Oflysosomes.

Ribosomes

Ribosomes Are Found In All Cells, Both Prokaryotes And Eukaryotes, Except In Mature Sperms And Rbcs. In Prokaryotic Cells, They Arc Found Floating Freely In

Fundamental-unit-of-life-ribosomesPNG

The Cytoplasm. In Eukaryotic Cells, They Occur Freely In The Cytoplasm As Well As Are Attached To The Outer Surface Of The Rough Endoplasmic Reticulum. They Are Also Found In The Mitochondria And Plastids.

Functions

Ribosomes Help In Protein Synthesis Inside The Cell Hence, They Arc Called Protein Factories Of The Cell.

Fundamental Unit Of Life KSEEB Chapter 1 Summary For Class 9 

Lysosomes

  • Lysosomes Are Membrane-Bound Sacs Filled With Digestive Enzymes. Ihese Enzymes Are Made By Rough Endoplasmic Reticulum. Lysosomes Have A Resistant Covering Membrane That Protects The Cell From The Digestive Enzymes Contained Inside Them.
  • Lysosomes Help In Waste Disposal From The Cell. Lysosomes Digest Any Foreign Material As Well As Worn- Out Cell Organelles, And Hence, Keep The Cell Free From Any Unwanted Waste Material. Whenever Any Foreign Material (Such As Bacteria) Enters The Cell, The Lysosomes Break Them Up Into Small Pieces. This Is Because The Enzymes Released By The Lysosomes Are Very Powerful And Help In The Process Of Digestion Of These Foreign Particles, Hi Case The Cell Gets Damaged, Lysosomes Burst And Release The Enzymes, Which Digest Their Own Cell. Hence, They Are Called Suicide Bags Of The Cell.

Functions

  • Lysosomes Help In Intracellular Digestion.
  • They Provide Energy During Starvation By Controlled Breakdown Of Stored Food.
  • Lysosomes Bring About Cellular Breakdown And Are Associated With Ageing.

Mitochondria – The Powerhouse Of The Cell
(Gk. Mitos: Thread, Chondrion: Granule)

Typically, The Mitochondria Are Sausage-Shaped Cell Organelles Covered By A Double-Membrane Envelope. The Outer Membrane Is Smooth And Porous, While The Inner Membrane Is Folded Into Structure Known As Cristae. These Folds Create A Large Surface Area For The Generation Of Atp (Adenosine Triphosphate) During Respiration. Mitochondria Release Energy In The Form Of Atp Molecules. This Energy Is Required For Performing Various Activities. Hence, Mitochondria Are Also Known As The Powerhouses Of Tire Cell Atp Is Known As The Energy Currency Of The Cell. Our Body Uses Energy Stored In Atp For Manufacturing New Chemical Compounds And For Mechanical Work.
The Cavity Of The Mitochondria Is Filled With Lipids, Proteins, Circular Dna And Ribosomes. Thus, Mitochondria Can Make Some Of Their Own Proteins.

Functions

Mitochondria Are Miniature Biochemical Factories Where Food Is Oxidized And Energy Is Released. This Energy Is Stored In The Form Of Atp. Hence, Mitochondria Are Called The Powerhouses Of The Cell.Fundamental-unit-of-life- mitochondriaThey Provide Important Intermediates For The Synthesis Of Several Biochemical Substances Like Chlorophyll, Cytochromes And Steroids.
Synthesis Of Many Amino Acids Occurs In Mitochondria.

Mitochondria Can Manufacture Some Of Their Own Proteins.
Plastids – Kitchen Of The Cell

Plastids Are Cell Organelles Found Only In A Plant Cell. Structurally, Plastids Contain Many Membrane Layers Embedded In A Material Called The Stroma. Like The Mitochondria, Plastids Also Have Their Own Dna And Ribosomes.
There Are Three Types Of Plastids – Chromo plasts (Coloured Plastids), Chloroplasts And Leucoplasts (White Or Colour less Plastids).
Chromo plasts (Gk. Chroma: Colour, Plasto: Formed) They Contain Fat Soluble Yellow, Orange Or Red Coloured Pigment. They Provide Colour To Flowers And Fruits. Chromoplasts Arc Formed Either From Leucoplasts Or Chloroplasts.

Chloroplasts (Gk. Chloros: Green,Plasto: Formed) They Are Green Plastids. Inside, Each Chloroplast Lias A Colourless Ground Matrix, Stroma And

Fundamental-unit-of-life- sub-micro-scopic-structure-of-the-chloroplast

A Membranous System, Grana. Each Granum Has Membrane-Bound Sacs Called Thylakoids. Thylakoids Of Chloroplasts Possess Photosynthctic Green Pigment, I.E. Chlorophyll. They Also Contain Various Yellow Or Orange Pigments In Addition To Chlorophyll.

Leucoplasts (Gk. Leukos: White, Plasto: Formed) These Are Colourless Plastids And Are Named On The Basis Of The Substances They Store. They Store Starch, Oils And Protein Granules.

Functions

  • Chromoplasts Arc Coloured Plastids That Provide Colour To The Flowers And Fruits.
  • Chloroplasts, The Green Plastids, Help In Photosynthesis And Thus, Help In The Synthesis Of
  • Food. These Are Called Kitchen Of The Cell.
  • Leucoplasts Help In The Storage Of Food.

Vacuoles

  • These Are Fluid-Tilled Membrane-Bound Spaces. Vacuoles Arc Storage Sacs For Liquid Or Solid Contents. They Are Bound By A Membrane Known As Tonoplast
  • In Animal Cells, Vacuoles Are Small. In Mature Plant Cells, The Small Vacuoles Fuse To Form A Single Large Central Vacuole. Hie Central Vacuole Of Some Plant Cells May Occupy 50-90% Of The Cell Volume.
  • In Plant Cells, Vacuoles Are Full Of Cell Sap And Provide Turgidity And Rigidity To The Cell. The Cell Sap Consists Of Tree Water And A Variety Ot Compounds In Solution Or Suspension. The Compounds Include Amino Adds, Minerals, Sugars, Organic Adds And Some Proteins.
  • In Singlc-Ccllcd Organisms Such As Amoeba, The Vacuoles Are Modified As Food Vacuoles. The Food Vacuole Contains Food Items That Amoeba Iras Consumed. In Some Other Unicellular Organisms, Like Paramecium, These Vacuoles Are Spedalized To Expel Excess Water And Some Wastes From The Cell.

Functions

  • Vacuoles Help The Plant Cells To Remain Turgid.
  • They Play An Important Role In Growth By Helping In The Elongation Of Cells.

difference-between-animal-cell-plant-cell

 

  • They Provide An Aqueous Environment For Die Accumulation And Storage Of Water-Soluble Compounds (Sugars, Minerals, Pigments, Etc.).
  • In Protozoans Like Amoeba And Paramecium, Vacuoles Help In Digestion And Excretion.

Centrosome

Centrosome Is A Small, Naked Protoplasmic Structure Present Near The Nucleus. It Is Present Only In Animal Cells. Centrosome Consists Of Two Small Granules Called Centrioles, Which Lie At Right Angles To Each Other. During Cell Division, Centrioles Migrate To Die Opposite Poles Of The Cell.

Functions

  • They Help In Spindle Formation During Cell Division. They Act As Basal Bodies And Give Rise To Cilia And Flagella.
  • Differences Between An Animal Cell And A Plant Cell
  • The Differences Between An Animal Cell And A Plant Cell Are Given

Cell Division

  • All New Cells Arise From Pre-Existing Cells By The Process Of Cell Division. New Cells Are Required For Growth, Replacement Of Old And Worn Out Cells And To Form Gametes During Reproduction. Cell Division Is Primarily Of Two Kinds Mitosis And Meiosis.

Mitosis

  • Mitosis Or Mitotic Cell Division Is An Equational Division In Which One Parent Cell Divides To Form Two Daughter Cells. The Daughter Cells Are Identical To Each Other And Also To The Parent Cell In Every Respect.
  • In Mitosis, The Same Normal Chromosome Number Of The Parent Cell Is Maintained At Each Stage Of Mitotic Division Of The Cell.

Meiosis

  • Meiosis Takes Place In The Reproductive Cells That Produce Gametes, Sperms And Ova. Meiosis Is A Modified Mitosis In Which Chromosomes Divide Once And The Nucleus Divides Twice. As A Result Of Which The Number Of Chromosomes Is Reduced To Half. Hence Meiosis Is A Reductional Division.

 

Fundamental-unit-of-life-different-stages-of-meiosis-in-animal-cell

Fundamental-unit-of-life-different-stages-of-meiosis-in-an-animal-cell

Summary

  • A cell is the structural and functional unit of life.
  • The cell was discovered by Robert Hooke in 1665 while he was studying a thin slice of cork under a self-built microscope.
  • The cell theory was formulated by two biologists, MJ Schlelden (1838) and T Schwann (1839).
  • On the basis of nuclear organization, cells are of two types – prokaryotic cells and eukaryotic cells.
  • A cell is enclosed by a plasma membrane which is made up of lipids and proteins. Plasma membrane is a living membrane. It is selectively permeable and allows only selected substances to pass through it.
  • In plant cells, a cell wall is also present. It is mainly composed of cellulose and is located outside the cell membrane. Cell wall provides a definite shape to the cell. It protects plasma membrane and internal structures from pathogens and mechanical injury.
  • In eukaryotes, the nucleus is separated from the cytoplasm by a double-layered membrane. It controls all metabolic and other activities of the cell. Hence, it is called the master of the cell.
  • Endoplasmic reticulum helps in intracellular transport. Hence, it is known as the circulatory system of the cell. It helps in the synthesis and transport of proteins and fats.
  • Ribosomes help in protein synthesis inside the cell. Hence, they are called protein factories of the cell.
  • Golgi apparatus consists of a system of membrane-bound vesicles which are arranged somewhat parallel to each other in stacks called cisterns.
  • Golgi apparatus helps in the storage, modification and packaging of substances manufactured in cell.
  • Mitochondria are miniature biochemical factories, where food is oxidized and energy is released. The energy is stored in the form of ATP. Plastids are cell organelles found only in plant cells.
  • Plastids are of three types – chromoplasts, chloroplasts and leucoplasts. Chloroplasts help in photosynthesis, and thus in the synthesis of food. These are called kitchen of the cell. Lysosomes help in Intracellular digestion. They also bring about cellular breakdown, and hence are called suicide bags of the cell.
  • In animal cells, vacuoles are usually absent. If present, they are small and scattered. But in mature plant cells, usually a single large central vacuole is present. It helps in maintaining the turgidity of the cell and stores important substances including waste.

Part A – Our PASTS – III (History)

Part B – Resources and Development (Geography)

Part C: Social and Political Life -III (Civics)

KSEEB Class 9 SSLC Biology Chapter 6 Improvement In Food Resources Notes

KSEEB Class 9 SSLC Biology Chapter 6 Improvement In Food Resources Notes Learning Objectives

  • After completing this chapter, you will be able to:
  •    recognize the necessity for regularly improving food production from agriculture and animal husbandry and need for sustaining them.
  • describe various methods of crop variety improvement such as hybridization and Introduction of genes into the crop;
  •  explain various methods of crop production management;
  •  list various methods of irrigating crop fields;
  •  describe rainwater harvesting technique:
  •  describe various cropping patterns such as mixed cropping, Intercropping, crop rotation, etc.;
  •  explain various means of crop protection management;
  •  differentiate between manure and fertilizers and list their various types;
  •  define animal husbandry and its importance;
  •  describe the methods of fish production, poultry farming and bee-keeping.

 animals and plants need food for their survival is the ultimate source of energy for all living beings. Different types of food crops are found in India .

  •  Food provides us with many nutrients like proteins, carbohydrates, fats, vitamins, minerals as well as water.
  • Food protects our body against diseases.
  • It is required for growth, development and repair of worn-out tissues of our body.

Since plants can manufacture their own food by photosynthesis, they are the ultimate source of food on the earth. However, for human beings, both plants and animals are the major sources of food. Mostly we obtain food from agriculture and animal husbandry. Agriculture (Latin: ager means field, and cultura means cultivation) is the cultivation (growing) of food crops in the field for food requirements. On the other hand, rearing animals (livestock) to obtain milk, eggs and meat and their management are part of animal husbandry.

KSEEB Class 9 SSLC Biology Chapter 6 Improvement In Food Resources Notes

Reasons For Imorovements In Agriculture and Animal Husbandry

Why is it necessary to regularly improve the food production from agriculture and animal husbandry? Why are the current levels of food production not enough? A few thousand years ago, most of the humans on the earth probably obtained their food by gathering leaves, fruits or roots and hunting animals.

improvementinfoodresourcesdifferenttypesofcropesfoundinindia

During that period, the human population was very low and there were enough food resources available in terms of animals, fruits, etc. However, if we see today, the scenario has totally changed and now India is a very populous country.

Currently, our population is more than one and a quarter billion people and is still growing. As population is increasing, the need for food requirement is also increasing. For this current population in India, we will need more than a quarter of a billion tonnes of foodgrains every year.

To get such a large quantity of foodgrains, we will need more land for farming. But, we know that the land for agriculture is limited and almost all available land is already extensively cultivated. Thus, there is not much scope to increase the area for farming and cultivation.

Hence, it is necessary to efficiently utilize the existing farming land and increase the efficiency of production of both crops and livestock. With new discoveries in science and technology and efforts in improving crop and livestock management, there has been some success in increasing food production. Our country has had two significant revolutions in agriculture and animal husbandry-green revolution and white revolution. Green revolution led to increased foodgrain (mainly wheat) production.

Likewise, white revolution led to better availability and more efficient use of milk in India. However, these revolutions caused over-exploitation of our natural resources. Due to more intensive use of land for crop production and subsequent loss of soil fertility due to over-exploitation of soil, land is becoming scarce, quantitatively as well as qualitatively. As a result, there are chances of damaging and destroying the balance of our natural resources.

By disturbing our natural resources, we will compromise with the productivity of agriculture. Therefore, there is a need for sustainable practices in agriculture and animal husbandry to increase food production without further depleting natural resources and disturbing the balance in nature. However, simply increasing food production cannot solve our problem of malnutrition and hunger. We know that India is largely an agriculture-based society as agriculture is one of the main sources of livelihood for majority of people in India.

Thus, it is necessary that the people engaged in agriculture get optimum returns from it in terms of income to buy food for combating the problem of hunger and malnutrition. For sustainable livelihood (healthy living), it is necessary to adopt sustainable agricultural practices such as mixed farming, intercropping, crop rotation, crop selection, integrated agriculture, etc.

Combining agriculture with poultry farming, fish culture, bee-keeping, livestock production, etc. is known as integrated agriculture. In this chapter, you will learn about different food crops, how they are cultivated and safely stored, animal husbandry and how to improve their production.

Improvement In Food Resources Chapter Summary Class 9 

Imorovement In Crop Yields

Different types of crops

Crops can be divided into two categories based on their utilization -food crops and fodder crops.

Food crops

Food crops include those crops that we eat to obtain nutrients. There are many types of food crops such as cereals, pulses, oil seeds, etc. In addition, there are vegetables, fruits and spices.
Cereals: Wheat, rice, maize, millets, barley and sorghum are cereals which provide us with carbohydrates that meet the energy requirements of our body.

improvementinfoodresourcesriceandwheatarecereals

 

improvementinfoodresourcesGroundnut,mustard,sunflowerandsoyabeanseedsaregoodsourcesofoil.

Pulses: Gram (chana), black gram (urad), pea (matar), green gram (moong), lentil (masoor), pigeon pea (arhar) are pulses and provide us with proteins. Oil seeds: Oil seeds include soyabean, groundnut, sesame (til), mustard, castor, sunflower and linseed.

Oil seeds provide us with oil which is a good source of necessary fats.
Vegetables, fruits and spices provide us with vitamins and minerals. In addition, they also provide us with small amounts of proteins, fats and carbohydrates.

 improvementsinfoodfoodresources

improvementinfoodresourcesvegetablesandfruitsprovidesuswithvitaminsandminerals

Fodder crops

Fodder crops are those crops which are raised as a food for the livestock. Fodder crops include berseem, oats or Sudan grass, etc.

Rabi Crops And Kharif Crops

For the growth and completion of their life cycle, different crops require different climatic conditions, temperature of the surroundings and duration of sunlight received, ie. photoperiod. Growth of plants and flowering depend on sunlight. Depending upon the season in which they are grown,
there are mainly two types of crops – rabi and kharif crops.

Rabi crops

The winter season is from November to April and is known as rabi season. Wheat, gram, peas, linseed and mustard are important rabi crops.

Kharif crops

The rainy season is from the month of June to October and is known as kharif season. The chief kharif crops are the millets, known as bajra and jowar, paddy, maize, cotton, green gram and black gram.

Activities For Improving Crop Yeild

If you compare the data of food production from 1952 to 2010, you will find that there has been an increase of 400 per cent (i.e. four times) in the production of foodgrains. However, the cultivable land area has increased by only 25 per cent during this period. How this increase in food production has been achieved? This increase in food production has been achieved by improving agricultural practices.

There are broadly three stages of farming:

  •  selection of seeds for planting,
  •  nurturing of crop plants, and
  •  protection of the growing crops and harvested crops from loss.
    The various tasks performed by a farmer to produce a good crop are called agricultural tasks or agricultural practices. Agricultural practices require systematic planning and proper management at all stages of crop production, starting from the selection of seeds and preparation of soil to harvesting and storage of food. Adopting good farming practices and managing resources like soil and water efficiently are the pre-requisites for this.
    The major activities for improving crop yields are given below.
  •  Crop variety improvement
  • Crop production management
  •  Crop protection management

Crop Variety Improvement

Need for crop variety improvement

A good variety of crop is one which gives better yield in minimum time, limited resources and optimum land utilization.

Improvement in variety means to develop varieties with desired characters such as higher yields, better qualities, resistance to diseases and pests, etc. It is mainly done by improving the genetic make-up of crop plants through plant breeding.

Plant breeding can be defined as the science of improving genetic make-up of plants in relation to their economic use. Varieties of plants can be selected by breeding plants for their various useful characteristics such as disease resistance, response to fertilizers, quality of product and high yield. Some commonly used plant breeding methods for improvement of crops/varieties are introduction of gene into the crop, hybridization, etc.

Hybridization

Hybridization is the process of crossing two genetically dissimilar (different) plants to obtain a progeny with the desired traits. This crossing may be:

  •  Intervarietal: Cross between two different varieties of the same plant species. This helps in producing high-yielding varieties.
  •  Interspecific: Cross between two different species of the same genus. This helps in producing plant varieties resistant to diseases.
  • Intergeneric: Cross between different genera.

Introducing a gene In this technique, a gene that would provide the desired characteristics is introduced in the plant. As a result, genetically modified crops are produced.

Characteristics of a good plant variety

  •  The new varieties of crops should produce high yields under different climatic conditions found in different geographical areas.
  •  The seeds should be of good quality and all seeds should be of the same variety and germinate under same conditions.
  •  Varieties should be capable of growing in diverse climatic conditions since weather conditions are unpredictable. This is also because cultivation practices and crop yields are related to weather, soil quality and water availability.
  •  The new varieties should also be tolerant to high soil salinity.

KSEEB Class 9 Biology Important Questions Chapter 6 

Objectives of crop variety improvement

Some objectives of crop variety improvement are asfollows:

  •  High yield: This is done to increase the productivity of the crop per acre. This can be brought about by developing high-yielding varieties by cross- breeding and hybridization.
  •  Improved quality: As human beings have become health conscious, they now stress on the quality of food with enhanced taste, nutrient value, etc. Quality considerations of crop products vary from crop to crop. For example, baking quality is important in wheat, protein quality in pulses, oil quality in oil seeds, and in fruits and vegetables, preserving quality is important.
  •  Resistance to undesirable biotic and abiotic stresses: New varieties developed as a result of varietal improvements should be resistant to biotic stresses like diseases, insects and other pests, and to abiotic stresses like drought, salinity, waterlogging, heat, cold and frost.
    Change in maturity duration: The shorter duration of crops from sowing to harvesting is more economical for the farmer. Maturity period of some long duration crops can be reduced, and then these crops can be used along with short duration crops in mixed cropping or multiple cropping. This will also reduce the cost of crop production. Uniform maturity will also make the harvesting process easy and it reduces the losses during harvesting. Using short duration crops, farmers can grow multiple types of crops in a year/season.
  •  Wider adaptability to new regions and climatic conditions: Improved varieties should have wider adaptability which would help in stabilizing the crop production under different regions and environmental conditions. Thus, single variety can be grown under different climatic conditions in different areas.
  • Desired agronomic characteristics: For fodder plants, tallness, high tillering and intensive branching are desirable agronomic traits, whereas dwarfness is the desired agronomic trait for cereal crops so that less nutrients are consumed by these cereal crops. Therefore, developing varieties of desired agronomic characters helps in getting more productivity from the same crop plant grown in a limited land area.
  • Varieties for saline soil: We must develop varieties which can grow equally well in saline soil and give normal returns. It will increase the production. The improved varieties of some important crops are given in Table 6.2.

improvementsinfoodresources

Crop Production Management

In India, like any other agriculture-based country, farming ranges from small to large farms. Different farming practices and agricultural technology are integral part of crop production management. There is direct correlation between economical condition of a farmer (input), his access to information and communication technology, and the yields he gets from farming. Thus, the cropping system and production practices depend upon farmers’ purchasing capacity.
Hence, there can be:

  •  no cost production practices,
  •  low cost production practices, and
  •  high cost production practices.
    Crop production management includes the following components:
  • Crop nutrient management
  •  Irrigation
  •  Cropping patterns

CROP NUTRIENT MANAGEMENT
Plants need certain chemical elements for their growth, development and metabolic activities. These chemical elements are called plant nutrients. There are about 30 to 40 elements found in plants, but only 16 of these are essential for their better growth and development. Other elements found in plant body are called non- essential elements.

The 16 elements found essential for growth and development of plants are – 1. Carbon, 2. Hydrogen, 3. Oxygen, 4. Nitrogen, 5. Phosphorus, 6. Magnesium, 7. Calcium, 8. Sulphur, 9. Potassium, 10. Manganese, 11. Iron, 12. Copper, 13. Zinc, 14. Boron, 15. Molybdenum and 16. Chlorine.

All elements found in plants are derived from air, water and soil.

  • Air: Carbon and oxygen
  •  Water: Hydrogen and oxygen
  •  Soil: Soil provides maximum number (13) of essential nutrients to the plant. The essential elements provided by soil are nitrogen, phosphorus, potassium, magnesium, sulphur, calcium, iron, manganese, boron, zinc, copper, molybdenum and chlorine.
    Although the soil provides 13 of the 16 essential nutrients required by plants, they constitute only about 0.5 to 6% of the plant tissues. These 13 elements provided by the soil are classified into two categories

improvementinfoodresourcessourcesofessentialplantsnutriunts

based on the quantity in which they are required by plants. These two categories are a. Macronutrients, and b. Micronutrients.

Macronutrients

Macronutrients are those elements which are utilized by plants in relatively large quantities. Out of the 13 soil nutrients, only six are macronutrients. These are nitrogen, phosphorus, potassium, calcium, magnesium and sulphur.

Micronutrients

Micronutrients are those elements which are required by plants in small quantities. Though required in small quantities, these nutrients are as essential for the growth and development of plants as the macronutrients. There are seven micronutrients required by plants. These are iron, manganese, boron, copper, molybdenum, zinc and chlorine.
The major differences between macronutrients and micronutrients are given in Table 6.3.

Deficiency of plant nutrients may affect the physiological processes in plants such as reproduction, growth and ability to fight against diseases. As a result, plants may show retarded growth, wilted leaves, low crop-yield, small seed kernel, increased susceptibility to diseases, etc. These nutrients can be added to the soil in the form of manure and fertilizers to get higher crop yield.

 

 

improvementinfoodresources

Manure And Fertilizers

Repeated cultivation of the same crops on the same agricultural land depletes the mineral nutrients and other useful contents of the soil. Due to this, the fertility of the soil is reduced, resulting in decreased crop production and lower economic yield. The fertility of an agricultural land can be enhanced by adding manure and fertilizers to the soil.

The necessary plant nutrients present in manure and fertilizers make the soil fertile. Nitrogen, phosphorus and potassium are the three most important elements for the healthy growth and development of crop plants, which can be supplemented in the soil through manure and fertilizer.

Manure

Manure is an organic substance obtained from the decomposition of plant and animal wastes like cow dung and plant residues. Manure is the major source of organic matter, which supply nutrients in small quantities but organic matter in large quantities and increase the fertility of soil. Manure contain a mixture of various nutrients recycled from biomass wastes (plant waste and animal excreta). Since long, the farmers in our country have traditionally used cow dung manure for replenishing soil nutrients.

Advantages of manure

Manure affect the soil in the following ways:

  •  Manure restore the soil texture for better retention of water and aeration, specially that of sandy soil. Hence, they improve the physical conditions of the soil.

 

improvementinfoodresourcescowdungmanure

  • They enrich the soil with nutrients. Since manure contain nutrients in small quantities, they are needed to be applied in large amounts in the field. They replenish a small part of the plant nutrients utilized by the crops.
  • They maintain the humus content of the soil.
  •  They add large amount of organic matter to the soil, which increases water-holding capacity in sandy soil and drainage in clayey soil.
  • Organic manure provide food for soil organisms, which help in making nutrients available to the plants.
  •  They are biological waste materials, which is advantageous as these protect our environment from harmful effects of excessive use of fertilizers. By using biological waste materials, farm waste is also recycled.

KSEEB SSLC Chapter 6 Notes Detailed Explanation 

Types of manure

On the basis of biological waste materials used, manure is classified into farmyard manure (FYM), compost, vermicompost and green manure.

Farmyard manure (FYM)

Farmyard manure (brown manure) is called so since it is prepared by using decomposed mixture of cattle excreta (dung) and urine along with the litter (generally straw) and leftover organic matter such as roughage or fodder. It is prepared by using the above- mentioned waste materials and storing them in a pit for decomposition by microbes till they form dark brown morphous substance, i.e. the manure.

Farmyard manure takes about 4 to 5 months for its formation (i.e. decomposition of waste material). A well-decomposed farmyard manure stimulates plant growth and seed germination. Residue of gobar gas plant is a type of farmyard manure.

Compost

Compost is prepared from farm and town refuse like vegetable waste, livestock excreta (cow dung), animal refuse, sewage waste, eradicated weeds, crop stubble and straw. This organic matter is decomposed by both aerobic and anaerobic microorganisms in pits. This process is called composting. Compost takes about 3 to 6 months for its formation. It is rich in organic matter and nutrients.

improvementinfoodresourcesacompostpit

Vermicompost

Compost is also prepared with the help of earthworms. Composting by introducing earthworms in the pit of organic matter is called vermicomposting. In this process, earthworms are introduced to the pit containing plant and animal refuse. Earthworms hasten the process of decomposition of plant and animal refuse.

Green manure

Green manure is prepared from young, fast growing green leguminous crop plants, about two months old, by ploughing them back into the soil. This is a practice of growing green plants prior to sowing of crop seeds and then ploughing of green plants into soil for the purpose of improving physical structure as well as soil fertility. The plants grown for the purpose of green maanuring are sunn hemp (Crotalaria juncea), dhaincha (Sesbania aculeata), guar (Cyamopsis tetragonoloba), berseem and cow pea.

These plants enrich the soil in nitrogen, phosphorus and other organic compounds. Generally, green manure crops like dhaincha and sunn hemp are grown in the fields of crops like rice, maize, sugar cane, cotton and wheat which require high nutrient input.

Limitations of manure

Manure is used extensively by farmers for the growth of crop plants. However, there are some limitations of manure.

  • Manure is relatively bulky material. Therefore, it is difficult to store and transport it.
  •  Manure can supply nutrients only to a limited extent and cannot replenish the massive depletion of nutrients caused by high-yielding varieties of staple crops.
  •  Manure release the nutrients slowly.
  •  Manure is not used to treat any specific mineral deficiency as it is not nutrient specific.

Fertilizers

Manure cannot supply all the essential plant nutrients to the soil. Therefore, they are to be supplemented with some chemical fertilizers. Fertilizers are the sources of plant nutrients manufactured commercially from chemicals. Fertilizers make the soil fertile. They are generally inorganic materials or salt containing nitrogen, phosphorus or potassium in the form of soluble chemical compounds.

Advantages of using fertilizers

  • Fertilizers are used to ensure high vegetative growth

 

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(leaves, branches and flowers) in order to produce healthy plants.

  • They are a factor in the higher yields of high cost farming.
  •  They are required in small quantity and hence can be easily stored and transported.
  •  Chemical fertilizers are nutrient specific that provide nitrogen, phosphorus and potassium as per requirement.

Chemical fertilizers, being soluble in water, are readily absorbed by the crops. Thus, chemical fertilizers are different from manure as manure is not soluble in water. The important differences between fertilizers and manure are given in Table 6.4.

Applications of fertilizers

Fertilizers contain much higher amounts of nutrients compared to manure, and therefore, should be used in very small quantities. The fertilizers are to be applied at a specific time of the plant growth cycle. The fertilizers can be applied before sowing, during irrigation, or sprayed on standing crops. High doses of these chemicals do increase crop yield, but these chemicals have many hazards also.

Disadvantages of using fertilizers

Important disadvantages of using chemical fertilizers are as follows:

  • These chemicals get washed away due to excessive irrigation and are thus not fully absorbed by the plants. These fertilizers reach rivers, lakes and other waterbodies and pollute them, disturbing the natural ecosystem.
  •  The continuous use of chemical fertilizers can cause drastic alterations in soil chemistry and affect the crop yield. Fertilizers also harm the soil microorganisms and thus destroy soil fertility as the organic matter in the soil is not replenished.
  •  The excessive use of nitrogenous fertilizers makes underground water rich in nitrates, which makes the water unfit for drinking,
  •  The chemical fertilizers, specially nitrogenous, can increase the biological oxygen demand (BOD) of water. This causes harm to aquatic animals like fish and aquatic plants. This phenomenon is known as eutrophication.
    Thus, for maintaining soil fertility and optimum yields, short-term benefits of using fertilizers and long- term benefits of using manure should be considered.

Organic Farming

Excessive use of fertilizers affects soil salinity as well as causes water and soil pollution. This is affecting our health. Therefore, nowadays, there is new system of farming which is used to produce food and fibre with minimal or no use of chemicals and with the maximum use of bio-agents and organic manure, known as organic farming.

In organic farming, there is maximum use of organic manure, recycled farm-wastes (such as straw and livestock excreta), and bio-agents (such as blue-green algae) in the preparation of biofertilizers with healthy cropping systems (mixed cropping, intercropping and crop rotation). Neem leaves or turmeric specifically are also used for grain storage as bio-pesticides. Organic farmers do not use synthetic pesticides or fertilizers.

Organic farmers use healthy soils by nourishing the soil with bio-agents that release, transform and transfer nutrients and green manure. Soil organic matter contributes to good soil structure and water- holding capacity. These produce healthy plants that are able to resist disease and insect predation better. In organic farming, healthy cropping systems such as mixed cropping, intercropping and crop rotation are used which effectively disrupt habitat for weeds, insects and disease organisms.

Advantages of organic farming

  • Toxin-free food: Organic food is safe to consume. Organic produce contains no or significantly lower levels of pesticide residues (which are already available in soil and water) than conventional produce.
  •  Controls pests and weeds: The cropping systems are beneficial in insect, pest and weed control besides providing nutrients.
  •  Environment friendly: Organic farming is environment friendly and does not cause air, water or soil pollution.
  •  Recycling of waste: It helps in recycling of wastes produced in the farm.
  •  Maintains fertility of soil: It helps in maintaining the fertility of the soil and soil structure.

Irrigation

Water is one of the most essential requirements for crop production. In India, more than half of our cultivable area does not have irrigation facilities and depends on rainfall for crop production. Thus, agriculture in India is rainfed and the success of crop yields in most areas depends on timely monsoon and sufficient rainfall during growing season.

However, the timing and extent of rainfall being uncertain, we cannot totally depend on rainwater for irrigating the crops. Rainfall may sometimes be excessive and cause floods, while other times, it may be scanty, leading to droughts. The efficiency of existing irrigation systems and the water supplied in fields can be increased with the judicious use.

We can improve the efficiency of irrigation by selecting appropriate crop and cropping system and using required quantity of water at appropriate time.
The process of supplying water to the crops by means of canals, reservoirs, ponds, lakes, wells, tube wells and tanks is known as irrigation.

Benefits of irrigation

  • Water supplies two essential elements to the crop plants, namely, hydrogen and oxygen.
  •  Seeds do not grow in dry soil. Irrigation provides moisture to the soil for the germination of seeds.
  • The roots of crop plants cannot grow well in dry soil. Irrigation loosens the soil and supports the growth and elongation of the roots.
  • Water at the time of irrigation dissolves the nutrients present in the soil. These nutrients dissolved in water are easily absorbed by the roots of the plants.

improvementinfoodresources

Types of irrigation system

The design, equipment and technique of replenishing the soil water by supplying irrigation water is known as irrigation system. Depending upon the kind of water resources available, several irrigation systems are adopted in India to supply water to the agricultural land. These include the wells, canal system, river lift system and tanks.

Wells

This system is based on the availability of ground water. In the well irrigation system, wells are constructed wherever usable groundwater is present. There are two types of wells, namely dug wells and tube wells. Dug wells collect water from water bearing strata and tube wells can draw water from deeper strata. From these wells, the water can be lifted by water pumps, transported through canals and used for irrigation.

The dug wells have their bottom below the groundwater table. The groundwater simply accumulates in the pit. From these pits, the water is lifted by mechanical means, either by using rope and container or bullock operated devices or by pumps. In tube wells, iron pipes are inserted deep in the ground through boring. The water through these pipes is lifted by diesel or electricity operated pumps. These tube wells can continuously supply water depending upon the need.

Improvement In Food Resources Class 9 KSEEB Question Answers 

Canal system

Canal system is an extensive and elaborate irrigation system. Big canals are made which receive water from water reservoirs or from rivers. The main canal is further distributed into branch canals and branch canals into field channels or distributaries. These distributaries may serve an individual field or a group of fields depending upon the area.

River lift system

In the river lift system, water is directly drawn from the rivers for irrigation in the areas near rivers. This method is adopted in those areas where canal flow is not sufficient or is irregular due to inadequate release of water from reservoir or river.

Tanks

Small storage reservoirs, which intercept and store the run-off water of smaller catchment areas are known as tanks. This technique is usually followed in the hilly areas having a slope. In this system of irrigation, small dams are constructed below the higher elevation of the catchment areas. From these dams or tanks, the outflow of water is regulated according to the availability of water, which helps in the uniform distribution of water.

River valley system

This system is operated in southern parts of our country, particularly in Karnataka and Kerala where many steep and narrow riverine valleys are found. During rainy season, there is high run-off and discharge flowing in these rivers. However, during rabi season, there is drying up of these riverines. In the valleys and the slopes of these narrow riverines, the crops like coconut, coffee, rubber and tapioca are cultivated.

Rainwater Harvesting And Watershed Management

Two fresh initiatives have been adopted for increasing the water availability for agriculture. These initiatives are:

  •  Rainwater harvesting
  •  Watershed management

Rainwater harvesting

In rainwater harvesting, the rainwater that falls on the

improvementinfoodresourcesrainwaterharvesting

ground or rooftops of buildings is channelled by canals and recharged into the ground by digging tunnels (Fig. 6.9a). This water percolates in the soil. Thus, the water table is maintained to a satisfactory level.

Rainwater can be stored in tanks

This is done by collecting water from cemented surfaces, such as rooftops in tanks or big containers (Fig. 6.9b). This water can be filtered and used in gardens to water plants, in fields for irrigation purposes, at home for cleaning vehicles, floor, clothes, dishes, etc. This water, when strained through a fine cloth, can be used as distilled water in batteries of vehicles.

Watershed management

It involves building of small check dams to increase percolation of water into the ground. It prevents rainwater from flowing away and also reduces soil erosion.

Cropping Patterns

In order to get maximum benefit, crops can be grown in different ways. Some of these ways have been described here.

Mixed cropping

Growing two or more crops simultaneously on the same piece of land is known as mixed cropping. Farmers have been growing two or more crops together on the same piece of land for many centuries. In India, different crops are used in mixed cropping.
Some of the commonly used mixed crops are:

  • Soya bean + Pigeon pea
  •  Sorghum + Pigeon pea
  •  Groundnut + Sunflower
  •  Maize + Urad bean
  •  Cotton + Mung bean
  •  Wheat + Gram
  •  Wheat + Mustard
    Criteria for the selection of crops in mixed
    cropping
    The selection of crops for mixed cropping is based on
    the following criteria:
  •  Duration of crops: One is short duration crop and the other is long duration crop. This helps in utilizing labour in a proper manner.
  •  Nature of crops: Food crops are usually mixed with cash crops to ensure both sustenance and income. Cereals and legumes are often mixed for the benefits of nitrogen-fixation by the legumes to the associated cereal crop.
  •  Root patterns: One crop may be deep-rooted whereas the other crop may be shallow-rooted.
  • Growth habits: One crop may be tall growing and the other may be short growing, so that these crops form different canopy.
  • Water needs: One crop may need more water and the other may need comparatively less water so that it may be used effectively.
  • Nutrient requirements: One crop may require more nutrients whereas the other may require less nutrients. This is done to reduce competition between member crops for water, nutrients and light. If one crop fails due to shortage of moisture or insufficient availability of nutrients, other crops can cover the risk of complete failure.

Advantages of mixed cropping

  • No risk of crop failure: Growing two crops of different nature simultaneously reduces the risk of total failure due to uncertain monsoon. This ensures some insurance against failure of the crops.
  • Increase in yield: Cereals and legumes are often mixed. Legumes have nodules having Rhizobium as nitrogen fixing bacteria. These bacteria fix up atmospheric nitrogen as nitrates in the soil. These nitrates make the soil rich in nitrogen. It benefits cereals or those crops requiring nitrogen as main nutrient. Thus, yield is increased.
  •  Minimizes pest damage: Mixed cropping reduces chances of pest infestation on any one of the two crops.
  •  Improves soil fertility: When a cereal crop (non- leguminous crop) is grown along with a leguminous crop, it helps in maintaining soil fertility. Legumes help in building soil fertility which gets depleted by growing cereal crops.
  • Less input of labour: Mixed farming involves greater flexibility of the distribution of labour and recovers the investment in much less time.

Intercropping

Intercropping can be defined as growing two or more crops simultaneously in the same field in a definite row pattern or tier system. In intercropping, the crops selected are those that have different nutrient requirements so that there is maximum utilization of nutrients available in the soil. In intercropping, a few rows of one crop alternate with a few rows of a second crop. For example, we can grow soya bean with maize, or pearl millet (bajra) with cowpea (lobia).

improvement infoodrsourcesintercropingofsoyabeanbymiaze

A good example is the multi-tier system of coconut, banana and pineapple or ginger or leguminous fodder or medicinal or aromatic plants. While ensuring biodiversity within a farm, intercropping also allows maximum use of resources.

Advantages of intercropping

Intercropping has several advantages over sole cropping:

  •  It makes optimum utilization of natural resources such as sunlight, land and water.
  • Intercropping also helps to prevent spread of pests and diseases to all the plants belonging to one crop in the field.
  •  It helps to maintain soil fertility.
  •  It economises space and time of cultivation of two or more crops.
    The comparison between mixed cropping and intercropping is given in Table 6.5.

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Crop rotation

When the same crop is grown in the same field, year after year, it removes particular nutrients from the soil and reduces its fertility. The fertility of the soil can be restored by crop rotation. Growing different crops on the same piece of land in a pre-planned succession is known as crop rotation.
In the rotation of crops, the legurainous crops like pulses, peas, beans, groundnut and Bengal gram are sown in-between the seasons of cereal crops like wheat, maize and pearl millet (bajra).

Crop rotation restores soil fertility

Cereal plants cannot utilize nitrogen directly from air, whereas leguminous crops possess the ability of fixing nitrogen from air to form nitrogenous compounds in the soil. In the root nodules of leguminous crops, the nitrogen fixing bacteria (Rhizobium) are present which can fix atmospheric nitrogen into nitrates. These are then utilized by plants for the synthesis of proteins.

 

improvementinfoodresourcesRootnodulesofleguminousplantcontainingnitrogenfixingbacteria

Thus, if a cereal crop is grown in a field, then most of the nitrogen present in the soil is absorbed and the soil becomes deficient in nitrogen. This reduces soil fertility. If the next crop grown is a leguminous crop, then it utilizes atmospheric nitrogen to form nitrogenous compounds in the soil. As a result, the deficiency of nitrogen containing compounds in the soil is removed. Hence, the soil becomes ready for growing the next crop.

Class 9 Biology Chapter 6 Agricultural Improvement Techniques 

Selection of crops for rotation

The selection of crops and their varieties is done on the basis of moisture conditions, length of the rainy season or irrigation facility, type of soil and risk involved.
Characteristics of crop rotation

  •  Legumes are grown in between cereal crops.
  •  Crops requiring high fertility level are grown after growing legumes.
  •  High input crops like wheat, rice, maize, sugar cane and potato may be grown before crops requiring low input.
  •  Generally, crops of the same family should not be grown repeatedly in the same piece of land as this will encourage growth of insects, pests and disease- causing organisms.
    Thus, while selecting crops for crop rotation, the
    following points should be kept in mind:
  • Availability of soil moisture through rain or irrigation
  •  Nutrient status of soil
  •  Availability of manure, fertilizers, pesticides, irrigation facilities, etc.
  •  Duration of the crop
    Depending upon the duration, crop rotation is
    classified as shown in Table 6.6.

Advantages of crop rotation

  •  Rotation of crops improves soil fertility. Hence, it brings about an increase in the production of foodgrains.
  • It saves a lot of nitrogenous fertilizers. This is because growing a leguminous crop during the rotation fixes atmospheric nitrogen and adds nitrogen to the soil at no cost.
  •  It helps in weed and pest control. This is because weeds and pests are very choosy about the host crop plant which they attack. When the crop is changed, the weeds and pests associated with that crop usually disappear.

Crop Protection Management

There are a large number of weeds, insect pests and diseases, which attack the food crops and damage them. Out of these, insect pests are one of the most destructive causative agents of field crops, which cause great losses in the agricultural produce. If these weeds, insect pests and diseases are not controlled at appropriate time, then they can damage the crop leading to a loss of more than 50 to 60% of its economic value.

Weed control

You would have seen that when we grow a crop, in addition to crop plants, certain unwanted plants also grow in the field simultaneously. These unwanted wild plants, which grow withimprovementinfoodresourcessomecommonweeds

crop plants in the cultivated field are called weeds. Like any other pest, they also adversely affect the crops by suppressing their growth. Weeds are very harmful because of the following reasons:

  • Weeds grow in the crop field and compete with the main crop. They consume a lot of soil nutrients, water, fertilizers, sunlight and space, which are meant for the crop plants. This lowers the quality of foodgrain (produce) and crop yield.
  •  Weeds may sometimes act as an alternate host for microorganisms and different insects, which feed on weeds and attack the crop plants, finally destroying them.

Common weeds

There are many types of weeds. The types of weeds vary from field to field, from crop to crop and also from season to season. Amaranthus (chaulai) is the most common weed, which grows with almost all crops. Xanthium (gokhroo), Parthenium (gajar ghas) and Cyperus rotundus (motha) are some more examples of commonly found weeds in India.

Methods of weed control

Weeds can be controlled by many methods, such as mechanical, cultural, biological and chemical.

Mechanical methods: Weeds are removed manually by uprooting them using hands, harrow (khurpi), hand hoeing, ploughing, land tilling, mowing (cutting with machines) and flooding.
Cultural methods: The cultural methods of weed control involve preparation of proper seed bed, timely sowing of crops, intercropping and crop rotation.

Biological methods: Biological methods of weed control involve the use of insects, which feed selectively on a particular weed. Certain microorganisms that cause diseases in the weed plants and eliminate them are also used in this method. For example, extensive growth of prickly pear weed (Opuntia) is checked through the introduction of Cochineal insect. Similarly, some crop plants like millet, barley and soyabean do not allow the growth of weeds. Such crop plants are known as smoother crops.

Chemical methods: There are certain chemicals that kill weeds. These chemicals are called weedicides. They are also known as herbicides. Some weedicides used to control different weeds are 2,4-D (2,4-Dichlorophenoxy acetic acid), Atrazines, Deltron, etc.

Insect pests

Any organism, which damages or destroys a crop plant is called a pest. Almost all crops are attacked by insect pests. The insect pests lead to low quality of foodgrains and cause a heavy economic loss to the farmer. They affect the health of the crop and reduce yield. Generally, insect pests attack the plants in three ways:

  • They cut roots, stem and leaves thereby destroying the crops.
  •  They suck the cell sap from various parts of the plant.
  • They bore into the stem and fruits and eat them from inside.

Control of insect pests

  • Insects infesting a food crop and other crop diseases can be controlled by various methods. The use of pesticides is one of the most common and effective methods of controlling insect pests and crop diseases. These pesticides are the chemicals used to control pests like insects, mites, rodents and fungi.
    There are following types of pesticides:
  • Weedicides: Chemicals used to control unwanted wild plants (weeds) in the crop field.
  •  Insecticides: Chemicals used to kill insects.
  •  Miticides: Chemicals used to kill mites.
  •  Rodenticides: Chemicals used to kill rodents
    (rats).
  •  Fungicides: Chemicals used to kill fungi.

Chemical methods of controlling insect pests

Those insects that harm the crop by cutting its root are controlled by mixing insecticides in soil. For example, chlorpyriphos is a chemical which is mixed in the soil to control root cutting type of insects.

Those insects that feed on stem and leaf by cutting them and also bore inside the fruits and grains are controlled by spraying insecticides like malathion, parathion and lindane on the crop plants. As the insects come in contact with these insecticides, the insecticides enter their body and kill the insects by affecting their internal metabolism. When an insect feeds on the plant sap, these insecticides enter inside the digestive system of the insect along with the cell sap and result in killing the insect.

However, we should not use pesticides in excess. Excessive use of a pesticide may create problems since they can be poisonous to plants and animals. They also cause environmental pollution.improvementinfoodresourcesmethodsofsprayingpesticides

Precautions for using pesticides

While using pesticides, following precautions should be taken:

  •  Pesticides should be used as per their recommended dose and specific instructions for applying them on crop field. A higher dose can be dangerous for the lives of human beings and other animals.
  •  The pesticides should be kept safely, well-packed and away from the reach of non-users like children.
  •  The people applying or spraying the pesticide on the crops should not come in direct contact with the pesticides as these are poisonous to human beings.

KSEEB SSLC Improvement In Food Resources Short Notes Class 9 

Crop diseases

mitted through water are called water-borne diseases. For example, blight of rice is a water-borne disease caused by bacteria.

A number of diseases attack crop plants. A plant disease is a structural or physiological abnormality which is injurious to the plant and reduces its economic value. Plant diseases are caused by microorganisms like fungi, bacteria and virus. Depending upon the mode of occurrence and transmission, plant diseases are classified into following three categories.

  • Soil-borne diseases: The diseases that get transmitted through the soil are called soil-borne diseases. For example, smut of bajra, tikka of groundnut and wilting of chick pea are some soil- borne diseases caused by fungi.
  • Air-borne diseases: The air also transmits many diseases. The diseases which are transmitted through air are called air-borne diseases. For example, rust of wheat is an air-borne disease caused by fungi.
  • Water-borne diseases: The diseases which are trans

improvementinfoodresourcessomecropdiseases

Preventive measures for crop protection

Chemicals (pesticides) are sprayed on crop plants, seeds and soil to kill the desired pests. However, we should try to avoid the use of chemicals as they cause environmental pollution. Instead, we should adopt the preventive measures rather than allowing the crops to be infested by pests and control them by using pesticides. Some of the preventive measures of pests are given below:

  • We should use resistant varieties of crops.
  •  The crops should be sown at their optimum time of sowing.
  •  We should adopt clean cultivation.
  • The fields should be ploughed during summer season to destroy weeds and pests.
  • There should be rotation of crops and multiple cropping system.

Storage Of Foodgrains

The foodgrains such as cereals and legumes are preserved for several months and sometimes even for a number of years. Therefore, it is necessary to

improvementinfoodresourcessomeinsectpestsoffoodgrains

keep them in safe storage to ensure their availability during all the seasons. Safe storage facilitates the distribution of food materials to distant places in the country. It also protects the foodgrains and other food materials from pests, rodents and other microorganisms.

Factors affecting stored food material

The various factors responsible for damaging the stored food materials are:

Abiotic factors

  • Inappropriate moisture content present in foodgrains and at the place of storage.
  •  Inappropriate temperature at the place where food material is stored.
  •  Humidity in the air.
  •  Properties of material of container in which food material is stored.

Biotic factors

  •  Insects, microorganisms (fungi, bacteria) and mites.
  •  Rodents, birds and other animals.
  • Enzymatic action within food material.

Methods of foodgrain storage

There are two different methods of food storage:

  •  Dry storage, and
  •  cold storage
  •  Dry storage: The storage of food materials at room temperature is called dry storage. In this method, all the non-perishable materials like foodgrains are stored.
  •  Cold storage: The storage of food materials at low temperature is called cold storage. The perishable food materials are stored in cold storage. The low temperature of cold storage reduces the losses from spoilage of food materials, and the nutritive value of the food materials is also retained for longer duration.

Preventive measures before foodgrain storage

Following preventive measures must be taken before storage of foodgrains for future use:

Cleaning of the produce before storage Before storage,

the harvested foodgrain should be checked to see if it contains any foreign matter like seeds, earth particles, sand and stone pieces, straw, etc. Any foreign material should be removed from foodgrains before storage.

Drying

Grains have a tendency to absorb moisture from the atmosphere. The moisture content of the grains has to be reduced to a certain level for safe storage of the grains. For drying, the foodgrains or seeds should be spread in thin layer on cemented floor or plastic sheets in the sun. After sun drying, the food materials are dried in the shade to allow them to cool to the room temperature before storing them in the containers, gunny bags or grain-silos.

Maintenance of hygiene in storage houses

Storage houses like godowns, stores, bins, etc. should be properly cleaned before storage of foodgrains in them. All sort of dirt, dust, rubbish, webbing or refuse of previous grain should be swept and removed. If there are any cracks or holes in the wall, floor or ceiling of the storage house, then these should be properly sealed.

Fumigation

Chemicals which can exist in gaseous state in sufficient concentration that can kill pests are known as fumigants. The use of fumigants to kill or repel pests is called fumigation. Before storage, the storage houses should be fumigated or treated with chemical sprays. Gunny bags should be treated with suitable insecticides before storage of foodgrains in them.

Animal Husbandary

In the previous sections, you have studied that agriculture evolved with the increase in population and food requirements to feed the growing population. As the population increased and the living standards also increased, the demand for milk, eggs and meat also increased. This necessitated the improvements in the livestock production.

Rearing animals (livestock) to obtain milk, eggs and meat and their management are called animal husbandry. In other words, the branch of agriculture related to the scientific management of animal livestock is known as animal husbandry. It includes breeding, feeding and disease control of domestic animals.

improvementinfoodresources

The various elements of animal husbandry are listed below:

  •  Proper feeding of animals
  •  Proper shelter for animals
  •  Proper care of animals against diseases
  •  Proper breeding of animals

Importance of animal husbandry

On the basis of utility, animal husbandry is important for the following reasons:

  • To increase milk production.
  •  To increase egg production.
  •  To increase meat production by rearing high meat-yielding animals.
  •  To increase production of edible and economically important fish.
  •  To properly utilize animal waste for economic purposes.
  •  To produce good quality, high-yielding breeds of domestic animals so that the production of food from animals is increased.
  • To properly manage domestic animals and benefit the farmer so that his income from them is increased.
  •  To increase the quality of food products from animals, thereby increasing the availability of nutritious food.

KSEEB Class 9 Biology Chapter 6 Notes 

Nutritional value of animal food

Animal food is generally rich in proteins. For example, milk and milk products are rich in protein content and are highly nutritious. Milk contains all the major food constituents like carbohydrates, proteins, fats, vitamins (A and D), water and minerals such as phosphorus and calcium. The major nutrients obtained from animal products are given in Table .

improvementinfoodresourcesIndigenousmichbreedsofcattle

 

Cattle Farming

We do cattle husbandry for two purposes:

  • For getting milk as a food.
  • For agricultural tasks such as tillage, irrigation and carting.
    On the basis of the above mentioned criteria, the animals can be divided into the following two categories:
  • Milk-yielding animals: These animals are also known as milch animals (dairy animals).
  • Dranght animals: These animals are used for farm labour.
    Indian cattle belong to two different species.
  •  Cows: Bos indicus, and
  • Buffaloes: Bos bubalis.
    The female cattle provide milk, which in turn contribute animal protein to the diet of people. The production of milk depends upon the duration of lactation period. Lactation period is the period of milk production after the birth of a calf. Thus, the milk production can be increased by increasing the lactation period.
    There are many important breeds of cows and buffaloes available in our country, which provide us

 

improvementinfoodrsourcesDifferentexcticbreedsofcows

good quality milk. The exotic or foreign breeds have long lactation period and provide us with good quantity of milk. The local breeds show excellent resistance to the diseases. We can also cross-breed the exotic breeds and local breeds to get animals with both the desired qualities (long lactation period and resistance to diseases).

Indigenous or local breeds of cattle: Sahiwal, Gir, Red Sindhi and Tharparkar are some of the indigenous breeds of cattle.

Exotic breeds of cattle: The exotic breeds mean those breeds which have been imported from abroad and are reared in India. The exotic breeds of cattle are Holstein-Friesian, Jersey, Brown Swiss, etc.
Improved breeds of cattle: In India, certain improved breeds of Indian cattle have been developed by making a cross between indigenous (local) cows with high milk-yielding with exotic breeds of bull. The improved, high milk-yielding breeds of cows developed in India are:

  • Frieswal
  •  Karan Swiss
  •  Karan Fries

Food requirements of dairy animals

Unlike human beings, animals cannot prepare their food. If they are domesticated, they must be fed fodder and grain. The animal food must be rich in nutrients for maintaining good health and for production of milk.

The food requirements of dairy animals is of the following two types:

  •  Maintenance requirements: These include food for normal metabolic activities required to support the animals to live a healthy life.
  •  Milk producing requirements: These include the food required during the lactation period.
    The animal food contains two types of substances: roughage and concentrates.
  •  Roughage are coarse and fibrous straw materials having a low nutrient content such as carbohydrates, fats, minerals, proteins and vitamins. Roughage form the bulk of the cattle feed. The animals get roughage in their feed from hay (straw of cereals) and fodder obtained from berseem, cowpea, husk of gram, wheat, rice, etc.
  •  The concentrates used in animal feed are mixture of substances, which are rich in protein and other nutrients but low in fibre. Concentrates include grains, seeds and oilseed cakes containing relatively smaller amounts of crude fibre and high amount of proteins and other nutrients. The cattle need a balanced ratio of all nutrients in appropriate amounts.
    In addition to roughage and concentrates, certain feed additives are also supplied in the diet of the cattle. The feed additives contain macronutrients and promote the health and milk output of the dairy animals.

Proper cleaning and shelter facilities

We must protect our animals from extreme environmental conditions (like heat, rain and cold). Thus, for rearing animals, we must provide them with proper shelters and clean them regularly.

Domestic animals should be regularly brushed to remove dirt and loose hair. They should be kept in well-lighted shelters with proper ventilation, water supply and hygienic disposal to protect them from rain, heat and cold. The floor of the rearing shelter should have proper slope so that water can drain easily and floor stays dry to facilitate cleaning. This will help in maintaining proper health of the animals and getting a good and clean milk yield.

Characteristics of a good animal shelter

An ideal animal shelter should have the following
characteristics:

  •  It should be situated at an ideal place with good surroundings.
  •  It should protect the animals from heat, cold, rain and extreme weather.
  •  It should be clean, dry and well-ventilated.
  •  It should get proper sunlight during the day and should be properly aerated.
  •  It should have proper arrangements and outlets for disposal of wastes including animal excreta.
  •  It should be spacious enough for each animal to stay comfortably and avoid overcrowding.

Fish Productio

Fish forms an important part of our diet. Fish is a valuable and cheap source of food rich in animal protein. Fish proteins are easily digestible. In addition, fish is useful for us in the following ways:

  • Medicinal use: Fish liver oil (Cod liver oil) is an important source of vitamin A and D.
  •  Industrial use: Body oil of some fish like herrings and sardines are used for the manufacturing of edible oil and margarine. Agricultural use: They are used as organic manure in the field.
  •  Feed for farm animals: Dried fish are used to provide proteins to farm animals.
  •  Adhesive: Skins and bones of fish are used to make high quality glues and adhesives.

Fish production includes fin fisheries [capture, management and exploitation of cartilaginous and bony fishes (true fishes)], and shell fisheries (capture, management and exploitation of crabs, prawns and molluscs).

improvementinfoodresourcesSomeexamplesoffreshwaterandmarineprawns

There are two ways of procuring fish:

  •  From natural resources, called capture fishery.
  •  By fish farming, called culture fishery. Fish grow either in seawater or freshwater, such as in rivers and ponds. Depending on the nature of the habitat (place where they are found), fishing can be done by both capture and culture methods in marine water and freshwater ecosystems. In addition, fish are also cultured in brackish water.
    Thus, fisheries are of following three types:
  •  Marine fisheries: This includes capturing fish of oceans and seas. Pomphret, tuna, sardine and mackerel are some marine fish.
  •  Freshwater fisheries: This includes capturing and culturing fish in freshwater system such as lakes, ponds, paddy fields, rivers, etc. For example, rohu, catla, Mystus and Gambusia are some varieties of freshwater fish.
  •  Brackish water fisheries: This includes fishing activities in brackish water (slightly salty water where sea water mixes with fresh water) such as lagoons, estuaries and mangrove swamps, etc. For example, pearl, spot and mullet are some brackish water fish.

Fisheries are places or establishments concerned with capturing, preserving, exploring and utilizing various types of fish, prawns, etc.

Improvement In Food Resources Chapter Summary Class 9 

Marine fisheries

Marine fishery resources in India include 7,516 km of coastline and the deep seas beyond it. In India, the main varieties of marine fish are pomphret, mackerel, tuna, sardine and Bombay duck. There are many types of fishing nets, hook lines, etc. which are used from fishing boats for capturing marine fish. Mechanical fishing boats and deep sea trawlers are now used to capture fish. Government

 

improvementinfoodresourcessomefish

As a result, fish production is increased within the same cost. Both local and imported fish species are used in composite fish culture systems. The fish species selected for composite fish culture have different feeding habits, and as a result, they do not compete for food among themselves. So, all the available food in the pond is effectively utilized. Catla, rohu, mrigal (Indian major carp) and silver carp, grass carp and common carp (exotic breeds of carp from China) are grown together.

Catla is surface feeder, rohu feeds in the middle-zone of the pond, mrigal and common carp are bottom feeders and grass carp feeds on weeds. Thus, together these species can use all the food available in the pond without competing with each other, leading to increase in the fish yield to great extent.

However, composite fish culture has many problems too. A major problem is the lack of availability of good quality seeds. Many of these fish breed only during monsoon. Even if fish seed is collected from the wild, it can be mixed with that of other species as well. Thus, to overcome this problem, these fishes are bred in ponds using hormonal stimulation. This has helped in increasing the supply of pure fish seed in desired quantities.

Activity 1

Request your teacher to organize an educational tour to a fish culture farm. Collect the following data from the fish culture farm.

  • Type of culture – monoculture or composite culture.
  •  Type of pond
  •  Varieties of fish in the ponds
  •  Feed ingredients supplied as fish feed in the fish farm
  •  What is the annual production capacity of the fish farm?
  •  What are the various diseases encountered by fish farmers?
    On the basis of your observations, prepare a project
    report and submit it to your teacher.

Poultry Farming

Poultry includes birds such as chicken (hen), duck, geese and turkey. However, it is mostly chicken that is domesticated on a large scale. Poultry farming is defined as the rearing and keeping of poultry birds such as fowl for eggs and meat. Poultry farming is done to raise domestic fowl for the production of eggs and chicken meat. Eggs and meat obtained from birds are a good source of nutrients. Therefore, improved poultry birds are developed and reared in farms to produce layers for eggs and broilers for meat.

The egg-laying poultry birds are called hen (egg layer) while the one groomed for obtaining meat is called chicken or broiler.

improvementinfoodresourcesSomebreedsofpoultrybirds

 

Variety improvement in poultry birds

Indian breeds of poultry birds provide good quality meat but lay small-sized and less number of eggs. Advantages of the desi breeds of poultry birds are that they are strong and have natural immunity against diseases in comparison to exotic breeds. However, exotic breeds are good as they produce more number of eggs and the size of eggs is also bigger than the desi ones.

Some of the indigenous (desi), exotic and upgraded breeds of poultry birds are mentioned here.

Indigenous (desi) breeds of poultry birds

1. Aseel, 2. Basara, 3. Chittagong, 4. Ghagus, s. Brahma, 6. Cochin

Exotic (Foreign) breeds of poultry birds

1. White Leghorn, 2. Rhode Island Red, 3. Black Minorca, 4. Plymouth, 5. Light Sussex

Upgraded breeds of poultry birds

Some improved varieties of poultry birds have been developed by cross-breeding the indigenous (e.g. Aseel) and foreign (e.g. Leghorn) breeds of hen. Some of the improved, high-yielding varieties of poultry birds are B-77, ILS-82 and HH-260.

The cross breeding programme for variety improvement to develop new varieties is focussed on the following:

  • To obtain large number and good quality of chicks.
  • To obtain dwarf (small) broiler parent for production of chick commercially.
  •  To obtain varieties which have capacity to adapt in summers and tolerance to high temperature.
  •  To get varieties which have low maintenance requirements.
  • To get the small-sized, egg-laying poultry birds which can easily utilize more fibrous cheaper diets formulated by using agricultural by-products.

Egg and broiler production

Poultry farming involves taking good care of birds for food, shelter and disease control.

In order to obtain good quality chicken for meat, broiler chickens are fed on vitamin-rich food for good growth rate and better feed efficiency. Special care is taken to avoid any mortality and maintain superior quality. Broiler chickens are produced as broilers and then sent to market for meat purposes.

It is necessary to maintain optimum temperature and hygiene in housing and poultry feed. In addition, it is necessary to control diseases and pests infestation. The housing (shelter), feed and environmental requirements of broilers are different from egg layers. The broilers should be provided with feed rich in proteins with adequate fat. The levels of vitamin A and K are kept high in poultry feeds.

Disease control

Since they live in crowded conditions, the poultry birds suffer from a number of diseases. These diseases are caused by bacteria, viruses, fungi, parasites and also from nutritional deficiencies. These diseases can be controlled by taking preventive measures like proper cleaning, sanitation, spraying of disinfectants at regular intervals, proper feed and vaccination. Appropriate vaccination is required to prevent any occurrence of diseases and loss of poultry birds in case of an outbreak.

Activity 2

Enquire about a nearby poultry farm and visit it. Note the types of breeds, food supplement given to poultry birds, and also the housing and lighting facilities provided there. Identify the growers, layers and broilers in them.

Bee-Keeping

Bee-keepingis the practice of rearing and taking care of honey bees on a large scale to obtain honey from them. It is also known as apiculture.
There are many advantages of bee-keeping:

  •  It provides honey, which is a nutritive food and has many other uses.
  •  It provides bees-wax, which is used in various medicinal preparations.
  •  Honey bees are excellent pollinating agents. Thus, they help in getting good agricultural yields.
    Thus, bee-keeping for making honey and other related benefits has become an agricultural enterprise. Since bee-keeping needs low investments, farmers use it as an additional way of income generation. Bee-farms or apiaries are established for commercial honey production.

Varieties of honey bees

There are four common varieties of honey bees. Out of these, three are local varieties while one is an exotic variety.

improvementinfoodresourcesHoneycombBee-keepingpractice

Indigenous (local) varieties

The indigenous (local) varieties of honey bees used for commercial honey production are:

  •  Apis cerana indica- commonly known as the Indian bee.
  •  Apis dorsata-commonly known as the rock bee.
  •  Apis florea – commonly known as the little bee.

Exotic variety reared in India

  • Apis mellifera (the Italian bee): This variety has been brought in India to increase yield of honey.

improvementinfoodresourcesArtificialbeehive

 

It can be easily domesticated and is commonly used for commercial honey production. This variety of bee has high capacity of honey collection. It also stings lesser than other varieties. It stays in a beehive for a long period and breeds very well.

Quality and taste of honey

The quality or value of honey depends upon pasturage or flowers available to the honey bees for nectar and pollen collection. + The taste of honey depends upon the quantity of pasturage and kind of flowers available.

Summary

  •  Cultivation of food crops in field for food requirements is known as agriculture.
  •  There are about 30 to 40 elements found in plants. Out of these, only 16 nutrients are essential for plant nutrition.
  •  The nutrients required by plants in large quantities are known as macronutrients whereas nutrients needed in small amounts are known as micronutrients.
  •  Soil provides nitrogen, phosphorus, potassium, manganese, magnesium, boron, iron, sulphur, copper, calcium and molybdenum to plants.
  •  Manure is an organic substance obtained from the decomposition of vegetable and animal waste, which supply  essential elements and humus to the soil and make it fertile.
  •  There are three types of manure namely, farmyard manure (FYM), compost and green manure.
  • Fertilizers are the plant nutrients manufactured commercially from chemicals.
  •  Fertilizers can be applied before sowing, during irrigation or spraying on the standing crops.
  •  A farming system with no or minimal use of chemicals and with maximum use of organic manures, etc. with healthy cropping system is known as organic farming.
  •  The various systems of irrigation used in India are: canal system, tanks, wells, river valley system and river lift system.
  •  Mixed cropping is growing two or more crops simultaneously on the same plece of land.
  •  Intercropping is growing two or more crops simultaneously on the same piece of land in a definite row pattern.
  •  Growing different crops on the same plece of land in a pre-planned succession is known as crop rotation. Crop rotation restores soil fertility.
  •  Improvement in variety simply means to develop varieties with desired characters such as higher yields, better qualities and resistance to different stresses, both biotic and abiotic.
  •  Weeds, insect pests and diseases can cause 50 to 70% damage to the crop. Insects, pests and weeds can be controlled by using pesticides and weedicides.
  •  The diseases of crop plants can be classified into soil-borne diseases, air-borne diseases and water-borne diseases.
  •  There are many pests of stored grains which damage stored food products. The stored grain pests can be controlled by chemicals, fumigation and plant products.
  •  We should use airtight, moisture impervious, thermally Insulated and rodent proof structures for safe storage of grains.
  •  The branch of agriculture related with scientific management of animal livestock is known as animal husbandry.
  • Animal husbandry helps in producing good quality, high-yielding breeds of domestic animals.
  •  Fish production includes capture, management and exploitation of fin and shell fisheries.
  •  Poultry farming is rearing and keeping of poultry birds for eggs and meat.
  • ee-keeping is the practice of rearing and taking care of honey bees on a large scale to obtain honey from them.

Key Terms

  • Agriculture: The cultivation of food crops in the field for food requirements
  • Sustainable agriculture: Successful management of resources for agriculture to satisfy the changing human
    needs, while maintaining or enhancing the quality of environment and conserving natural resources
  •  Food crops: Crops that we eat to obtain nutrients
  •  Fodder crops: Crops that are raised as food for the livestock
  •  Rabi crops: The crops that are raised in the winter season i.e. from November to April
  •  Kharif crops: The crops that are raised in the rainy season i.e. from June to October
  •  Plant breeding: The science of improving genetic make-up of plants in relation to their economic use
  •  Hybridization: Process of crossing two genetically dissimilar plants to obtain a progeny with the desired traits
  •  Intervarietal: Crossing between different varieties
  •  Interspecific: Crossing between different species of the same genus
  •  Intergeneric: Crossing between different genera
  •  Macronutrients: Those elements which are utilized by plants in relatively large quantities
  •  Micronutrients: Those elements which are required by plants in small quantities
  •  Weedicides: The chemicals which are used to kill weeds
  •  Pest: Any organism which damages or destroys a crop plant
  •  Dry storage: The storage of food material at room temperature
  •  Cold storage: The storage of food materials at low temperature
  •  Fumigation: The use of fumigants to kill or repel pests
  •  Milch animals: Milk-yielding animals
  • Draught animals: The animals that are used for farm labour
  •  Lactation period: The period of milk production after the birth of a calf
  •  Mariculture: Culture of marine fishes
  •  Aquaculture: Rearing and management of aquatic animals
  •  Apiculture: The practice of rearing and taking care of honey bees on a large scale to obtain honey from them

KSEEB Class 9 SSLC Biology Chapter 3 Diversity in Living Organisms Notes

KSEEB Class 9 SSLC Biology Chapter 3 Diversity In Living Organisms Learning Objectives

  • After completing this chapter, you will be able to: establish a relationship between classification and evolution
  • recognize the hierarchy in classification of groups
  • understand the Importance of binomial nomenclature over common names of plants and animals
  • classify living world into five kingdoms and give theirrharartArierire;
  • classify kingdom Plantae Into various divisions; classify kingdom Animalia into phyla and give characteristics and examples of each phylum; classify phylum Chordata Into classes, giving their characteristic features and examples.

Have You Ever Visited A Zoo?

  •  In a zoo, you find variety of animals. Similarly, if you look in a garden or a park, you will find several types of plants – some with flowers, some without flowers, some shrubs, some large trees and so on. If you compare yourself with, a monkey, you will find several differences as well as many similarities too. Did you know that there are about 10 million species of organisms on earth? However, only one-third of them have been identified so for.
  • There are different life forms on the earth ranging from microscopic bacteria to gigantic blue whale of length 30 metres and redwood tree of California of 100 metre height. There is variety in lifespan as well as form and colour of organisms. If we attempt to study such a vast variety of organisms, it will be convenient if we group them and give each of them a biological name.
  • In this chapter, you will study about diversity in the living world, classify living world into five kingdoms and establish a relationship between classification and evolution.

Biodiversity

What Is Biodiversity?

Biodiversity means the diversity of life forms or occurrence of different living organisms and their variety in a particular region. The term biodiversity was coined by Walter G. Rosen in 1985.

Class 9 KSEEB Biology Chapter 3 Diversity in Living Organisms Study Material 

KSEEB Class 9 SSLC Biology Chapter 3 Diversity in Living Organisms Notes

Does biodiversity remain unchanged over a period of time?

  • Since in any given area many different life forms live together and share tire environment and its resources, they arc affected by each other also. During this process of co-existence, a stable community of different species is formed.
  • In recent times, the balance of community with environment has also largely changed due to human involvement The diversity in such communities is affected by characteristics of land, water, climate, etc. Changes in the genetic make-up also bring about morphological changes leading to biodiversity.

What are the regions of rich biodiversity?

  • The tropical regions ranging between the Tropic of Cancer and the Tropic of Capricorn are rich in diversity of microorganisms, plant and animal life.
  • This is because of the warm and humid climate in the tropical region. This region is also called the region of mega diversity. Only a few countries contain more than half the biodiversity available on earth.
  • These are Brazil, Colombia, Peru, Ecuador, Mexico, Zaire, Madagascar, Australia, China, India, Indonesia and Malaysia.

What is Classification

  • Classification means identifying similarities and differences between different kinds of organisms and then placing these groups in an order of close relationship.
  • In other words, biological classification attempts to group living organisms according to how closely related these organisms are. This is called phylogenetic system of classification.In the eighteenth century Carolus Linnaeus considered as ‘Father of Taxonomy’ developed a way of naming and classifying organisms.
  • The study of the kinds and diversity of organisms and the evolutionary relationships among them is known as systematic.? (Gfc. systema – system, ikos – body of facts) or taxonomy (Gfc. taxis – arrangement, L. nominalis – belonging to a genome). These studies result in the description of new species and the organization of animals into groups based on evolutionary relationships.
  • To understand classification in a simple way, let us take one example. Suppose you are given a basket full of different kinds of fruits, and you are asked to separate them on the basis of their similarity and morphological appearance. Then you will place oranges with oranges, grapes with grapes, apples with apples and so on.

Diversity-in-Living-Organisms-a-simple-diageam-to-show-classification.

What is the Basis of Classification?

  • Since time immemorial, attempts have been made to classify living things into groups. Aristotle classified animals according to very simple way by just looking at life as to whether they lived on land, in water or in air. However, this categorization did not provide an exact basis of classification as there were differences among animals living in water, on land or in air. For example, in water you find many lands of fish, corals, octopus, etc. All of them live in water, yet they are very different from each other.
  • Thus, it is necessary to first identify characteristics on the basis of which the broadest division can be made. Then, this is followed by the next set of characteristics for making sub-groups within these broader divisions. In this way, each time a new characteristic can be taken to further classify each group.

What do we mean by characteristic?

  • A characteristic Is a particular form or function exhibited by organisms. For example, five fingers in each hand in humans is a characteristic. Plants do not contain fingers and thus are different from humans.
  • There are some characteristics which are broader and widely found. On the other hand, there are some characteristics that arc at microscopic level. The broader characteristic is followed by limiting characteristics or small characteristics.
  • The characteristic at the next level would be dependent on the previous characteristic and would decide the variety in the next level.
  • This way a hierarchy of mutually related characteristics, to be used for classification, is built.

Some examples of characteristics used for a hierarchical classification are

  • Presence or absence of membrane-bound organelles: All eukaryotic cells have membrane- bound organelles such as nucleus and mitochondria. The membrane bound organelles allow cellular processes to be carried out efficiently in isolation from each other. In organisms that do not have dearly demarcated nucleus and other organelles, biochemical pathways take place in a very different way. This would have an effect on ever)’ aspect of cell structure. Thus, presence or absence of membrane- bound organelles will act as a basic characteristic in classification of different organisms.
  • Occurrence of cells singly or in groups: This is another characteristic that is taken into consideration while classifying living beings. Do cells live singly (unicellular) or in groups (multicellular) and does each group live as an individual group? Some cells such as Amoeba live singly, while some cells group together to form an individual. Cells that group together to form a single organism use the principle of division of labour.
  • In such cases, all the cells forming an individual would not be similar. Instead, group of cells would cany out specialized functions. This would result in a very basic distinction in the body design of organisms. As a result, you find many differences in the body structure of single-celled organisms such as Amoeba and multicellular organisms such as worms and humans.
  • Mode of food procurement: What is the mode of procurement of food? Whether organisms produce their own food through photosynthesis (autotrophs) or depend on food manufactured by others (heterotrophs)?
  • What is the level of organization in the photosynthetic organisms (plants)?
  • In animals, how does an individual’s body develop and organize its various parts. What is the specialization in organs to perform different functions?
  • In all of the above questions, a hierarchy is developing. Tims, the characteristics of body structure used for classifying plants will be different from those used for classifying animals as the basic structures of plants and animals arc different These structural features are then further used to make sub-groups, rather than making broad groups.

KSEEB Biology Chapter 3 Diversity In Living Organisms solved Questions

Diversity in Living Organisms Importance of classification

  • A wide variety of organisms on the earth can be easily studied by classification.
  • Classification gives a summary of all life forms at a glance.
  • It helps us understand the relationship between different groups of organisms and helps in understanding their evolutionary history.
  • It serves as a base for the development of other biological sciences such as ecology and behavioural sciences, biogeography and evolution.
  • It helps in understanding the characteristics of whole group of organisms by studying only a few representatives from that group.
  • It helps us in the identification and classification of different organisms such as agricultural pests, pathogens and components of an ecosystem on which various fields of applied sciences such as agriculture, public health and environmental biology depend.

Diversity in Living Organisms Classification And Evolution

  • In the previous section, you have studied that all living things are identified and categorized on the basis of some characteristics such as their body structure in form and function, mode of nutrition etc.
  • Out of many characteristics, some characteristics are dominant and are likely to produce wide-ranging changes in body structure than others. Time also plays an important role in this. Some characteristics appear earlier in the body structure while other characteristics appear late. Thus, those characteristics which came into existence earlier arc likely to be more basic and fundamental than characteristics that came into existence later. So, we can say that once a certain body structure has come into existence, it will be the basis for other changes in the body structure that happen subsequently. Thus, the classification of life forms will be closely related to their evolution.
  • However, one question arises here as to what is evolution?
  • Evolution is the process of gradual change or development in different kinds of living organism. Most life forms that exist today have arisen by an accumulation of changes in body structure over a period of time. This has allowed an organism to possess them for his/her better survival. It was Charles Darwin who for the first time described this idea of evolution in 1859 in his book, The Origin of Species.
  • This idea of evolution can be linked to classification. According to Darwin, there are some organisms with primitive body structure that have not changed much over a period of time. These are called older or primitive or lower organisms, which arc simple. On the other hand, there are other groups of organisms that have acquired their peculiar body structure quite recently. They have a complex body structure. They may be considered as the younger or advanced or higher organisms. Since the complexity in body structure is likely to increase over a period of time, we may say that older organisms are simpler, while younger organisms arc more complex and have evolved from the simpler organisms.

KSEEB SSLC Chapter 3 Diversity In Living Organisms Key Points For Class 9 

The Hierarchy In Classification Of Groups

Each, organism is classified on the basis of its evolutionary relationship with other groups of organisms. Biologists such as Ernst Haeckel (1894), Robert Whittaker (1959), and Carl Woese (1977) tried to classify all living organisms into broad categories. These broad categories are called kingdoms.

Kingdoms – Two Or five ?

  • The largest group of organisms recognized by biologists is the kingdom. But how many kingdoms should be there? Earlier, there were only two kingdoms of classification, namely plants and animals. This, however, caused problems in classifying fungi, bacteria and single-celled organisms, which do not fit into either kingdom – plant or animal.
  • R H Whittaker in 1969, suggested the five kingdom classification on the basis of cell structure i.e. presence or absence of a nucleus, body organisation (unicellular or multicellular) and mode of nutrition. The five kingdoms that are recognized by Whittaker are -Monera, Proto ctista (Protista), Fungi, Plantae and Animalia. Later another Biologist, Woese, introduced modification by dividing Monera into Archacbactcria (or Archaea) and Enbacteria (or bacteria).
  • further classification is done by naming sub-groups at various levels in the following scheme

Diversity-in-Living-Organisms-hierarchy-of-classification.

  • By grouping organisms on the basis of hierarchy of characteristics into smaller groups, one can arrive at the basic unit of classificatiou which is species.

Species

  • The lowermost category is the species. Each organism is classified into a particular species. It is the basic unit of classification.
  • Members of a species interbreed to produce fertile offsprings, which can perpetuate.
  • Apart from small variations, members of a species arc almost identical in their anatomy, physiology and behaviour.
  • Members of a species often resemble each other very closely in appearance (but are not exactly similar). For example, tigers belong to species tigris, humans belong to species sapiens. Thus, the scientific name for tigers is Panthera tigris and for humans is Homo sapiens.

Genus

  • Closely-related species are grouped into a genus (plural genera). Thus, genus is a group of species which have common external resemblance. For example, lion (Panthera leo), leopard (Panthera pardus) and tiger (Panthera tigris) are three different species but belong to one genus Panthera.

Family

  • Groups of similar genera are grouped together into a family. For example, cat (Felis domestica) and tiger (Panthera tigris) belong to the same family, Felidae.

Order

  • Similar families are grouped together into orders. For example, humans and apes arc grouped together into the same order, Primates. Git (Felidae) and dog (Canidae) are grouped together into the same order, Carnivora.

Class

  • Similar orders arc grouped into classes. For example, we belong to the class Mammalia along with other orders like Carnivora that includes tiger, dog and cat.

Phylum

  • Closely-related classes arc grouped into phylum. For example, classes like mammals, fish, amphibians, reptiles and birds all belong to the phylum Chordata.

Kingdom

  • Similar phyla constitute a kingdom. For example, all animals belong to kingdom Animalia and all plants belong to kingdom Plantae.

KSEEB Class 9 Biology Chapter 3 Diversity In Living Organisms Explained 

Nomenclature-Scientific Naming of Organisms Activity 1

  • To study the common names of some animals and plants in different languages
  • Look for the names of some common animals and plants. Find out the common names of these in different languages.

 

Diversity-in-Living-Organisms-activity-1.

  • From the above activity, we can conclude diat different animals and plants have different names. A cat is called billi in Hindi, biral in Bengali and punai in Tamil. There are different names of a cat in Arabic, Russian and French.
  • Thus, a need was felt to assign a particular organism with an internationally acceptable name in the same manner as the chemical symbols and formulae of various substances are used throughout the world. Thus, a simplified system of naming organisms called Binomial nomenclature was proposed by Carl Linnaeus (1707 1778).
  • Binomial nomenclature means a two-name system of naming organisms. The scientific name of an organism is thus unique and can be used to identify it anywhere in the world.
  • Under Binomial system of nomenclature, each organism lias a scientific name consisting of two parts. The first part is genus and the second part is the species. For example, the scientific name of mango is Mcmgifera indica, scientific name of house fty is Musca domestica and scientific name of the common Indian frog is Rana tigrina

Mainly, the following rules arc applied to scientific names:

The name of the genus comes first and its first letter is always capitalized. The name of the species comes second and starts with a small letter. For example, the binomial name for humans is Homo sapiens. In this, Homo is the genus, starting with a capital letter while sapiens is the species starting with a small letter

  • The scientific name is printed in italics or underlined (separately for genus and species name) if handwritten. For example, Homo sapiens or
  • Scientific names are mostly in Latin and are accepted all over the world.

Nomenclature-Scientific Naming of Organisms Activity 2

  • To study the scientific names of some common plants and animals
  • Given below is a table giving common names of some plants and animals. Find out the scientific names of these and write them down

Diversity-in-Living-Organisms-activity-2.

Diversity-in-Living-Organisms-activity-2.-1

Five Kingdoms

Kingdom Monera

(Gk. monos: single)

  • The organisms belonging to kingdom Monera are the oldest, simplest and prokaryotic microorganisms. They have following characteristics:
  • These are prokaryotes. They do not have a defined nucleus or organelles.

Diversity-in-Living-Organisms-bacteria-and-blue-green-alge-kingdom-monera.

  • They are unicellular, but may be present in the form of colonies or filaments of independent cells.
  • Some of them have cell walls while others do not have (for example Mycoplasma).
  • Their mode of nutrition is mainly autotropluc (i.e. synthesizing their own food, for example some bacteria and blue-green algae) or sometimes heterotrophic (i.e. getting it from the environment by absorption, for example Mycoplasma).
  • They are usually non-motile, but some may have flagella and gliding movements. Cilia arc absent.
  • – Their reproduction is primarily asexual (by fission).
  • Examples: Bacteria (E. coliVibrio cholerat etc.), blue-green algae or cyanobacteria (Anabama, Nostoc etc.) and Mycoplasma.

Kingdom Protista

(Gk. Protista: first)

The organisms belonging to kingdom Protista include unicellular eukaryotic organisms. Primarily, they are aquatic and occur in oceans, lakes, ponds and damp soils.

Diversity-in-Living-Organisms-some-examples-of-protists

They have the following characteristics:

  • They are first eukaryotes, having a well-defined nucleus and complex membranous organelles. They are unicellular or colonial forms without distinct division of labour.
  • Some have cellulose wall and chloroplast and their mode of nutrition is autotrophic (algae). Some do not have cell wall and chloroplast and are heterotrophic (Protozoans),
  • They have hair-like cilia (Paramecium) or whip- likc flagella (Euglena) or pscudopodia (Amoeba) for moving around.
  • Examples: Unicellular algae, diatoms, protozoans (Paramecium, Amoeba, Euglena), etc.

Class 9 SSLC Biology Chapter 3 Notes With Diagrams 

Kingdom Fungi

  • The organisms belonging to kingdom Fungi include heterotrophic (without chlorophyll) eukaryotic organisms living as parasites or saprophytes.

 

Diversity-in-Living-Organisms-some-examples-of-fungi

  • They have a definite cell wall made up of tough complex sugar called chitin.
  • They grow in dark and moist habitat and use dead organic matter as food. Hence, they are called saprophytes.
  • Many of them have the capacity to become multicellular organisms at certain stages in their lives except yeast which is a unicellular fungus.

Diversity-in-Living-Organisms-the-five-kingdom-classification.

  • They arc usually non-motilc.
  • They reproduce mostly by spore formation. However, sexual reproduction may also take place. Examples: yeast, mushrooms, etc.
  • Some fungi live in mutual relationship with algae or blue- green algae (cyanobacteria). Such a relationship is called symbiotic relationship. These symbiotic forms are called lichens that are usually found growing on the bark of trees.

Kingdom Plantae

  • Kingdom Plantae includes all organisms that arc truly multicellular eukaryotes with cell walls. All plants are included in this kingdom. Die plantae are distinguished by the following characteristics:
  • They are multicellular eukaryotic organisms adapted to carry out photosynthesis. Hence, they are autotrophic.
  • They perform photosynthesis with the help of chlorophyll present in cliloroplasts.
  • They have a cell wall made up of true cellulose, enclosing cytoplasm with large vacuole.
  • They are non-motile, without definite shape or size.
  • We will stud)r about kingdom Plantae in later sections of this chapter.

Kingdom Animalia

  • Kingdom Animalia includes all those organisms which are multicellular eukaryotes without cell walls. They are heterotrophs as they caimot synthesize their own food. We will study about kingdom Animalia in later sections of this chapter.

Kingdom Plantae (Plant Kingdom)

  • All plants arc multicellular, photosynthetic and contain chloroplast in their cells. Most of the plants are adapted for living on land.
  • Plants can be classified at three levels.
  • Level one: Whether plant body is well- differentiated and has distinct components.
  • Level two: Whether the differentiated plant body has special tissues for the transport of wrater and other substances within it  Level three: Whether plants are able to bear seeds and whether the seeds are enclosed within the fruits.
  • The plant kingdom includes algae, mosses, liverworts, ferns, conifers and flowering plants.

Sub-KingdomCryptogamae

(Gk. crypto: hidden, gamous: marriage)

  • The plants belonging to sub-kingdom Cryptogamae are also known as lowrer group of plants, flow’erless or seedless plants. ’Ihesc plants do not bear external flowers or seeds and hence are considered to have hidden reproductive organs. The sub-kingdom cryptogamae is further divided into three divisions, namely, Thallophyta, Bryophyta and Pteridophyta.

Division Thallophyta

(Gk. Txatlus: undifferentiated, phyton: plant)

Characteristics of division Thallophyta

  • The plant body is not differentiated into roots, stems and leaves. It is in the form of an undivided mass of cells called thallus.

Diversity-in-Living-Organisms-classification-of-plants.

  • Vascular system is absent, o These plants are predominantly aquatic.
  • The thallophytcs mainly include algae.
  • Examples: Spirogyra, Chora, Cladophora, Ulothrix, Ulva, etc.

Division Bryophyta

(Gk. Bryotu moss, phyton: plant)

The bryophytes are amphibians of the plant kingdom. The bryophytes include mosses and liverworts. They usually range from 2 cm to 15 cm in length. They are unable to grow tall because they do not have xylem to support them, nor do they have a vascular system for transport of materials. They do not have true roots.

Diversity-in-Living-Organisms-some-examples-of-fthallophytes-algae

Characteristic features of division Bryophyta

  • These are simplest land plants with no true roots. The body of the plants is simple, flat and usually differentiated into stems and leaf-like structures.
  • They do not have specialized tissues for conduction of water and other substances horn one part of the plant to another. This means that bryophytes do not have conducting or vascular tissues,xylem and phloem. Thus, they are called noil-vascular cryptogamic plants.
  • The plants develop a number of hair-like structures at their lower ends. These are known as rhizoids. Rhizoids help in fixing the plant to the soil and absorbing nutrients.
  • Tire)’ haw chlorophyll and thus can manufacture their own food tlirough photosynthesis.
  • ‘ Bryophytes show alternation of generations, i.e. they possess two types of generations – gametophyte and sporophyte.
  • These plants mainly live in damp, shady places.

Diversity-in-Living-Organisms-some-examples-of-bryophytes

  • Examples: Ricciti, Mafchantia (liverwort), Funaria (moss), etc.

Division Pteridophyta

(Gk. Pteroti: feather, phyton: plant)

  • The term ‘pteridophyta is derived from the Greek word pteron meaning feather. This is because pteridophytes have fcathcr-likc or pinnate fronds. Pteridophytes are seedless vascular plants that have sporophytic plant body. They are also called vascular cryptogams as they possess well-developed conducting system, i.e. vascular elements such as xylem and phloem. They grow well in shady areas like forests, mountain slopes and hills, especially in the tropical rain forests.

Class 9 SSLC Biology Chapter 3 Notes With Diagrams

Characteristic features of division Pteridophyta

  • The body of pteridophytes is differentiated into true stem, leaves and roots like higher plants.
  • They possess well-developed vascular tissues – xylem and phloem – for conduction of water and other substances from one part of plant body to another.
  • Tire most conspicuous of pteridophytes are the ferns. They arc seedless vascular plants of humid tropical and temperate areas.
  • Examples: Ferns (Nephrolepis), horse-tails (Equisetum), club mosses (Selaginella) and Marsilea. The major differences between Bryophyta and Pteridophyta.

Diversity-in-Living-Organisms-some-examples-of-pteridophyta

Diversity-in-Living-Organisms-differences-between-bryophytes-pteridophyta

Sub-Kingdom Phanerogamae

(Gk. Phaneros: visible, gamos: marriage)

  • The plants belonging to sub-kingdom Phanerogamae arc 9ccd bearing plants. The body of these plants i3
  • differentiated into true roots, stem and leaves. ‘Ihe vascular system composed of xylem and phloem is well- developed. The reproductive organs are multicellular seeds and an embryo develops from a fertilized egg. The major differences between Cryptogamae and Phanerogamae are given.
  • Sub-kingdom Phanerogamae is further sub-divided on the basis of presence or absence of fruits (Le. whether the seeds are naked or enclosed in fruits) into two sub-divisions – Gyranospenoae and Angiospcrmae.

Sub-division Gymnospermae

  • The term gymnosperm is derived from two Greek words – gyntno meaning naked and sperma meaning seed. Thus, gymnosperms are plants with naked seeds. They are found mostly in cold areas where snowfall occurs too often. All gymnosperms are perennial woody plants. Some of these live for thousands of years.

Characteristics of Gymnospermae

  • The plants are differentiated into roots, stem and leaves.
  • The stem is erect. It may be branched a.3 in Pinus or unbranched as in Cycas.
  • Gymnosperms are a small group of seed plants. Seeds do not occur inside a fruit. They are naked or lie exposed on the surface of megasporophylls. o The seed contains a food-laden tissue called endosperm for the future growth of embryo.
  • Examples: Cycas, Pinus, Ginkgo, Araucaria, Gnetum etc.

Diversity-in-Living-Organisms-conifers-and-cycas-gymnospermae.

Diversity-in-Living-Organisms-differences-between-cryptogamae-phanerogamae

Diversity-in-Living-Organisms-differences-between-gymnogamae-angiospermae

Sub-division Angiospermae

  • The term angiosperm is derived from two Greek words – angio meaning covered and spertna meaning seed. Thus, angiosperms are plants with seeds covered by tire fruit ‘Die major differences between Gymnospermae and Angiospermae are given
  • On die basis of number of corylcdons, angiosperms are divided into two groups – monocotyledonous or monocots (seeds with a single cotyledon) and dicotyledonous or dicots (seeds with two cotyledons). The major differences between monocots and dicots are given .

Free Notes For KSEEB Class 9 Chapter 3 Diversity In Living Organisms

Characteristics of Angiospermae

  • These are flowering plants the seeds of which are enclosed in a fruit.
  • The seeds contain cotyledons called seed leaves as in many cases they emerge and become green when the seed germinates. Thus, cotyledons represent pre-designed plant in the seed.
  • After fertilization, ovary develops into a fruit.
  • Xylem contains vessels and phloem contains companion cells.
  • Examples; Pea, gram, maize, oats, rice, Hibiscus, Paphiopedilum, Ipomoen, mango, etc.

Diversity-in-Living-Organisms-some-examples-of-angiosperms

Diversity-in-Living-Organisms-differences-between-monocots-dicots

Activity-3

To study dicot and monocot seeds and plants Yon will need Seeds and plants of green gram, wheat, maize, peas, tamarind and water

Procedure

Take some seeds of green gram, wheat, peas, maize and tamarind. Soak them in water for one day till they become tender.

Try to split each seed into two halves. Do all of them split into nearly two halves?

Observation 1

  • ‘The seeds that split into rwo halves are dicot seeds and die seeds that do not split are monocor seeds.
  • Now take the plants of these and observe their roots, leaves and flowers.
  • What is the root system – tap roots or fibrous roots?
  • What is the type of venation – parallel or rericulate?
  • How many petals are there in the flowers of these plants?

Carefully observe the above-mentioned characteristics and note them down in a table in your workbook.

Observation 2

You will observe that the wheat and oiaize plants (monocot) have fibrous root system, parallel venation and three or multiple of three petals.

On the other hand, dicot plants like green gram, peas and tamarind have tap root system, reticulate venation and five or multiple of five petals.

Diversity-in-Living-Organisms-activity-3.

Kingdom Animalia

Let us discuss the various characteristic features of

kingdom Animalia before classifying them.

Characteristic features of kingdom Animalia

  • These are multicellular organisms.
  • These arc eukaryotic organisms, that is, the cells do not contain cell wall, but contain only cell membrane and membrane-bound organelles (like nucleus).
  • They are heterotrophs.
  • They are mostly mobile.
  • They show increased sensitivity through the nervous system.
Let us now study about the phyla included in kingdom animals

Major Phyla Of Kingdom Animalia

Based on die extent and type of body structure differentiation, the kingdom Animalia is divided into two groups, namely, Invertebrate (non-chordata) and Chordata. A broad division of kingdom Animalia is given in

Phylum Porifera – organisms with pores

‘Ihe word porifera means organisms with pores or holes all over their body. Phylum Porifera includes sponges. Porifera are multicellular organisms with cellular level of body organisation.

KSEEB Class 9 Diversity In Living Organisms Chapter 3 Notes With Examples

Characteristics of Phylum Porifera

  • These are non-motile organisms fixed to the solid surface. They are commonly called sponges and found in marine habitats, i.c. oceans.
  • The body has many pores (ostia) or holes. These pores form canal system. Water flows through canals and brings food and oxygen with it. It is then passed outside through osculum, a large aperture present at the upper end of the body.
  • Body encloses a large cavity called spougococl.
  • Body is covered with a hard outer skeleton.
  • The body of sponges has minimal cellular differentiation and almost no tissue organization. That means that though sponges arc multicellular organisms, their cells are not organized to form tissues.
  • They are mostly sessile. No organs, movable parts or appendages are present.
  • The sponges reproduce asexually by budding and gcnunulc formation and sexually tlirough fertilization. They have great regenerating capacity. Examples: Sycott, Euplectella, Easpongia, Spongilla, Leucosolaua, etc.

Diversity-in-Living-Organisms-kingdom-animalia

 

Diversity-in-Living-Organisms-classification-of-animals

Phylum Coelenterata (Cnidaria)

  • The phylum Coelenterata includes Hydra, jelly fish, sea anemones and corals. Coelenterates are multicellular organisms which have tissue level ot organization. Their body consists of two layers of tissues (ectoderm and endoderm) with a jelly-like substance in between.
  • The coelenterates mostly live in fresh water, in solitary or colonial forms. Some like Hydra may be seen with naked eyes and can be found attached to stones or aquatic plants.

Characteristics of Phylum Coelenterata

  • The body has no head and no segmentation. The body shows tissue level of organization.
  • These animals have two germ layers, namely, ectoderm and endoderm; one layer makes up the cells on the outer side of the body (epidermis)

Diversity-in-Living-Organisms-some-spong-belong-to-phylum-porifera

  • and other layer makes the inner lining of the body (gastrodermis). There is a jelly-like substance in between.
  • The mouth is surrounded by a ring of tentacles. Tentacles contain nematocysts (stinging cells) called cnidoblasts.
  • The body has a gut cavity (called coelenteron) with a single opening for food and waste material.

Diversity-in-Living-Organisms-some-examples-of-phylum-cndaria

  • Some of the species of coclcntcrates live in colonics (such as corals), while some live a solitary life (such as Hydra).
  • Reproduction may be asexual!)’ or sexually.
  • Examples: Hydra, Obelia, sea anemone (Metridiutu), Aurelia (Jelly fish), etc.
  • The major differences between Porifera and Coelenterata (Cnidaria) are given in Table 3.5.
  • Hydra. a coelenterate, has great regenerating power. If it is cut into several fragments, each fragment will grow into a new Hydra.

Phylum Platyhelminthes – the flatworms

(Gk. Platy: flat; helminth: worm)

  • The Platyhelminthes are flatworms. Flatworms live in moist places. The body of platyhelminthes has complex body organization in comparison to poriferans and coclcntcrates.
  • They range from a few millimetres to a few centimetres in size.

Characteristics of Phylum Platyhelminthes

  • Their bodies are dorsoventraUy flattened. That is why, they are called flatworms.
  • The body is bilaterally symmetrical with the left and right sides of the body having the same design.
  • They are triploblastic animals (body made of three germ layers – ectoderm, mesoderm and endoderm from which tissues can be formed). This allows formation of outer and inner bodylinings and some organs.
  • They possess a mouth but no anus.
  • There is some degree of tissue f ormation. However, there is no true coelom or internal body cavity, in which well-developed organs can be accommodated (housed).

Diversity-in-Living-Organisms-some-examples-of-phylum-platyhelminthes

Diversity-in-Living-Organisms-differences-between-porifera-and-coelenterata

  • They arc either free-living (such as planarians) living in water and soil or parasitic (such as liver flukes) living inside the body of other organisms. They have suckers or hooks or both for attachment to the body of the host.
  • Reproduction takes place mostly sexually, but few reproduce asexually too. They are mostly hermaphrodite (both sex organs in the same body). Examples: Tapeworm, liver fluke (Fasciola hepatica), Planaria, etc.

Phylum Nematoda (or Nemathelminthes)- Parasitic worms or roundworms

(Gk. Nema: thread; helminth: worm)

  • The organisms belonging to this phylum are also known as Aschelminthes. They are mostly parasites having slender, elongated bodies tapering at each end.

KSEEB SSLC Biology Chapter 3 revision Notes For Diversity In Living Organisms

Characteristics of Phylum Nematoda

  • They are triploblastic animals (body wall having three germ layers).
  • The body is bilaterally symmetrical and unsegmented, and is cylindrical (rather than flattened).
  • The body has tissue level of organization. However, no real organs are present.
  • The body cavity is a pseudocoelom, i.e. no true coelom is present
  • The organisms are mostly parasites having slender, elongated bodies tapering at each end.
  • They are parasitic worms that cause diseases such as elephantiasis (caused by filarial worms) and intestinal disorder (caused by roundworms or pinworms). Ascaris lumbricoides is a common roundworm living in the intestine of humans.
  • They have a definite digestive system that runs along the length of their body. They have a mouth, pharynx, intestine and anus. Sexes are separate, males are smaller than females.

Diversity-in-Living-Organisms-ascaris-and-wuchereria-belong-to-phylum-nemathelminthes

  • Examples: Ascaris(roundworm),Enterobius (pinworm), Wuchereria (filarial worm)etc.
  • The major differences between Platyhclminthes and Nemathelminthes are given.

Phylum Annelida – the ringed worms

(Gk. annulus: ring; idos: form)

  • Organisms belonging to phylum Annelida arc worms with body appearing as if made up of a series of rings. It includes earthworms, ragworms (Nereis) and leeches. They arc first animals with true coelom.

Characteristics of Phylum Annelida

  • The are elongated, segmented, worm-like animals.
  • Body is triploblastic and contains true coelom. This allows true organs to be packaged in the body.

Diversity-in-Living-Organisms-some-examples-of-phylum-annelida

Diversity-in-Living-Organisms-differences-between-plathelminthes-and-nemathelminthes

  • The body is bilaterally symmetrical.
  • The have organ level of organization, i.e. extensive organ differentiation.
  • They have a well developed brain, ventral nerve cord, dorsal and ventral blood vessels.
  • The body is provided with setae or parapodia for locomotion.
  • Their body is mctamcrically segmented (made up of a series of similar segments). Each segment is lined up one after the other from head to tail, and possesses a similar pattern of organs, such as nerves, blood vessels and excretory organs.
  • They are found in diverse habitats ranging from freshwater, marine water to land.
  • Examples: Pheretima (earthworm), leech, Nereis (sandworm), Aphrodila (sea mouse), etc.

Phylum Arthropoda – organisms with jointed appendages

(Gk. Arthro: jointed; pod: legs)

  • Organisms belonging to phylum Arthropoda have jointed appendages (legs and antennae).
  • This is the largest invertebrate phylum. Insects, centipedes, millipedes and spiders arc the common organisms belonging to phylum Arthropoda.

Characteristic features of Phylum Arthropoda

  • The body is bilaterally symmetrical and segmented.
  • They are triploblastic animals.
  • The body (coelomic) cavity is reduced and filled with blood, thus called haemocoel. There is an open blood circulatory system and the blood docs not flow in well-defined blood vessels.

Diversity-in-Living-Organisms-some-examples-of-phylum-arthropodaPNG

  • The body is segmented into two regions – cephalothorax (head and thorax together) and abdomen, or three regions – head, thorax and abdomen.
  • They have jointed appendages (legs).
  • They possess a hard exoskeleton made of chitin.
  • Examples: Crab, butterfly, dragonfly, housefly, prawn, scorpion, spider, etc.

The major differences between Annelida and Arthropoda are given

Diversity-in-Living-Organisms-differences-between-annelida-and-arthropodaPNG

Phylum Mollusca

  • The organisms belonging to phylum Mollusca have an outer shell and soft body. The fold of skin around the body is called mantle which secretes the outer shell. They have muscular foot which is used for movement.
  • Phylum Mollusca includes slugs, snails and octopus.

Characteristics of Phylum Mollusca

  • These are triploblastic animals with a coelom- The coelomic cavity is reduced.
  • The body is bilaterally symmetrical.
  • They live in terrestrial or aquatic habitat
  • The body is unsegmented or with little segmentation.
  • The body is soft, often protected by a hard calcareous shell.
  • They have a muscular foot on their ventral surface for moving around.
  • They have an open circulatory system.
  • They have kidney-like organs for excretion. Examples: Snail, Chiton, Pila, Utiio, octopus, etc

Diversity-in-Living-Organisms-some-examples-of-phylum-mollusca

Diversity-in-Living-Organisms-some-examples-of-phylum-echinodermata

 

Phylum Echinodermata – organisms with spiny skin

  • The word Echinodermata has come from Greek words echinos meaning spiny and derma meaning skin. Thus, they are organisms with spiny skin.
  • Characteristics of Phylum Echinodermata
  • They are triploblastic coelomate animals.
  • They are exclusively free-living marine animals living in sand at the bottom of the sea.
  • Their body is unsegmented. There is no head and body surface lias five radial anns (pentamerous symmetry).
  • They move by tube feet The body cavity has water-vascular system.
  • They possess calcareous exoskeleton made up of calcium carbonate with spiny skin all over the body.
  • They have a water driven tube system that they use for locomotion.
  • Examples: Asterias (starfish), Echinus (sea urchin), Holothuria (sea cucumber), Antedon (feather star), etc

Class 9 KSEEB Biology Chapter 3 Taxonomy And Classification Notes

Sub-Phylum Protochordata

The word Protochordates means primitive chordates. They possess notochord at some stage of their life. Notochord (chord means string) is a stiff rod-like structure that runs along the back of the animal and separates nervous tissue from the gut It also provides a place for muscle attachment for easy movement.

Diversity-in-Living-Organisms-some-examples-of-sub-phylum-protochordata

Characteristics of Sub-Phylum Protochordata

  • They are triploblastic animals with a coelom.
  • The notochord may not be present in all stages in their lives. It does not run along the entire length of die animal body.
  • They are marine animals.
  • Examples: Batanogfossus, Herdmania, Amphioxus.

Phylum Chordata

(Sub-Phylum Vertebrata)

The vertebrates are triploblastic, coelomate animals with bilaterally symmetrical body. They include the fish, amphibians, reptiles, birds and mammals. Vertebrates have a true vertebral column and internal skeleton, which allows a different distribution of muscle attachment points for movement of the body parts. They have complex differentiation of body tissues and organs. The major differences between Nonchordata and Chordata are given in Table 3.8.

Characteristics of chordates

  • Presence of notochord: In some stages of their life, all chordates possess a stiff rod-like structure (called notochord) running along the body, close to die dorsal surface.
  • Dorsal tubular nerve cord: They have a dorsal, hollow, tubular nerve cord close to the notochord.
  • They are triploblastic animals with diree germ Layers.
  • They have paired gill pouches in some stages of their life.
  • They are true coelomate animals.
  • Vertebrates are fiirther grouped into five classes.

Group Pisces (fish)

The group called Pisces contains many classes. They are all exclusively water living (aquatic) animals. Some well-known examples are cartilaginous fish (entirely cartilaginous endoskeleton), such as sharks and bony fish (bony and cartilaginous endoskeleton) such as tuna or rohu. The cartilaginous fish belong to Class Chondrichthycs, and the bony Osh belong to Class Osteichthyes. These are the two classes under the Pisces group.

Characteristics of Group Pisces

  • They have a cartilaginous skeleton or bony skeleton.
  • They live exclusively in water. The body is streamlined, which allows easy movement in water. They possess fins and muscular tail for locomotion.
  • They possess gill slits for water to move inside die body for respiration. Pour to seven pairs of gills are present. They obtain oxygen dissolved in water with die help of gills.
  • Eyes are without eyelids. They possess nostrils that help in olfaction.

 

Diversity-in-Living-Organisms-differences-between-nonchordata-and-chordata

 

Diversity-in-Living-Organisms-some-examples-of-pisces

 

  • The body is covered with overlapping dermal scales. The)1 are cold-blooded animals (i.e. their body temperature changes with the change in the outside temperature).
  • The heart is two-chambered, unlike four-chambered heart in mammals.
  • They lay eggs.
  • Examples: Mandarin fish, angler fish, lion fish, rohu, sea horse, climbing perch, liilsa, mullet, electric ray, stingray, dog fish, etc.

Diversity-in-Living-Organisms-some-examples-of-class-amphibia

 

Class Amphibia

(Gk.Ainphi: double or both; bios: life)

The amphibians have evolved from fish. Thus, they are adapted to partly live on land, and partly in water. They differ from fish as they do not possess scales. They have three-chambered heart and mucous glands in die skin.

Characteristic features of Class Amphibia

They live partly in water and partly on land. Larval stage has a tail and live in water Adults may live in water as well as on land.

Diversity-in-Living-Organisms-external-features-of-frog.

Diversity-in-Living-Organisms-differences-between-amphibians-and-reptiles

  • They have smooth skin. It is covered by mucus, which keeps the skin moist and slippery.
  • They have two pairs of limbs. Digits do not have claws.
  • Respiration takes place either duough gills or lungs.  Heart is thrce-chambercd.
  • Most amphibians lay their eggs in water, where they develop into tadpoles. The tadpoles after metamorphosis become adults and can live in water as well as on land.
  • Examples: Toad, frog, salamander, tree frog, etc

Class Reptilia (Gk. Repere: to crawl)

They are the cold-blooded animals with their body covered with horny epidermal scales. The animals belonging to Class Reptilia can live solely on land and are very less dependent on water. They lay eggs covered with waterproof shells, so they need not lay eggs in water, like amphibians. Their skin is waterproof and covered with protective homy scales.

Characteristics of Class Reptilia

  • They are cold-blooded animals.
  • Their body has a head, neck, trunk and a tail.
  • Their body is dry and covered with homy, scaly epidermal skin.
  • The)- have two pairs of limbs on their trunk.
  • They have lungs tor breathing and respiration.
  • Most of them have three-chambered heart (four-chambered in crocodiles).
  • They lay eggs covered with waterproof shells, so they need not lay eggs in water like amphibians. Examples: Wall lizards, snakes, turtle, flying lizard,chameleon, crocodiles, etc.

Summary Of KSEEB Biology Chapter 3 Diversity In Living Organisms For Class 9

Class Aves (Gk. Avis: bird)

Birds evolved from reptiles and have many similarities with them. Birds are easily distinguished because their bodies are covered with feathers. All birds are warm¬blooded and have four-chambered heart.

Diversity-in-Living-Organisms-some-examples-of-class-reptailsPNG

Characteristics of Class Aves

  • They have feathers and scales covering their bodies.
  • They are warm-blooded animals (Le. their body temperature does not change with outside temperature).
  • o The body is divisible into three parts – head, neck and trunk.
  • Their forelirabs are modified to form wings tor flight.
  • Heart is four-chambered.
  • Their jaw has a homy beak- Teeth are absent.
  • They have lungs for respiration.
  • They lay eggs.
  • Examples; Pigeon (Columba livia), house sparrow, crow, white stork, penguin, male tufted duck, etc.

 

Diversity-in-Living-Organisms-some-examples-of-aves

Class Mammalia

(Gk. Mamma: breast)

  • All mammals possess mammary glands that are more developed in females than in males to suckle their young ones. Like birds, mammals are also warm¬blooded animals. Mammals have body covered with hair.
  • All female mammals secrete milk on which their young ones feed.

Characteristics of Class Mammalia

  • They possess mammary (milk) glands. They secrete milk to feed their young ones.
  • Their body is covered with hair. Sweat and oil glands arc also present in the skin.
  • The body is divided into head, neck, trunk and tail. Tail is absent in some mammals. They have four limbs.
  • Except two mammals,the platypus and the echidna(spiny ant-eater) which lay eggs,all other mammals give birth to young ones.Kangaroos give birth to very poorly developed young ones.
  • Sexes arc separate – male and female.Their young ones usually develop inside the uterus of die mother.
  • Heart is four-chambered.
  • They are warm-blooded animals.
  • Examples: Rat (Rattus rattus), human, tiger, elephant, cat, whale, bat, etc.

Diversity-in-Living-Organisms-some-examples-of-mammals

 

Diversity-in-Living-Organisms-differences-between-aves-and-mammals

Summary

  • Classification means grouping organisms on the basis of their similarities and dissimilarities.
  • The scientific naming of organisms is according to the Linnaeus system of Binomial Nomenclature.
  • As per modern Whittaker system of classification, there are five kingdoms of living organisms namely, Monera,Protista, Fungi, Plantae and Animalia.
  • Monerans do not have a defined nucleus or cell organelles, nor do they show multicellular body structure.
  • Protists include unicellular eukaryotic organisms such as algae, diatoms and protozoans.
  • Organisms belonging to kingdom fungi are heterotrophic eukaryotic organisms which use decaying organic matter as food material.
  • Yeast and mushrooms are Included in kingdom fungi.
  • All multicellular eukaryotes, which are photosynthetic autotrophs, are grouped under kingdom Plantae.
  • Kingdom Plantae is classified into two sub-kingdoms – Cryptogamae and Phanerogamae.
  • Thallophyta are plants that have undifferentiated body.
  • Bryophyta are amphibians of plant kingdom.
  • Pteridophyte are vascular plants that do not produce seeds.
  • Gymnosperms are plants that bear naked seeds.
  • Angiosperms are plants that bear seeds enclosed inside fruits.
  • Angiosperms are divided into two classes depending upon the number of cotyledons in their seeds monocotyledons and dicotyledons.

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