Class 11 Physics

KSEEB Class 11 Physics Solutions For Chapter 10 Mechanical Properties Of Solids Important Points

Elasticity: The property of a body by virtue of which it tends to regain its original size (or) shape when the applied force is removed.

Plasticity: It is the inability of a body not to regain its original size (or) shape when external force is removed.

Substances that exhibit plasticity are called plastic substances.

Stress (σ): The restoring force per unit area is called stress (σ).

Stress (σ) = Force/area Unit Nm^-2 (or) pascal; D.F: ML^-1T^-2

Tensile Stress: When the applied force is normal to the area of cross-section of the body then restoring force per unit area is called tensile stress.

Tangential (Or) Shearing Stress: The restoring force developed per unit area of cross-section when a tangential force is applied is known as shear stress or tangential stress.

Hydraulic Stress (Volumetric Stress): For a body in a fluid force is applied on it in all directions perpendicular to its surface.

“The restoring force developed in the body per unit surface area under hydraulic compression is called hydraulic stress.”

Strain: The change produced per unit dimension is called strain. It is a ratio.

Mechanical Properties Of Solids Notes In KSEEB Physics

Strain Types:

1. Longitudinal Strain: The ratio of increase in length to original length is called as longitudinal strain. Longitudinal strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\)
2. Tangential (Or) Shear Strain: The ratio of relative displacement of faces Δx to the perpendicular distance between the faces is called shear strain. Shear strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\) = tanθ
3. Volume Strain: The ratio of change in volume ΔV to the original volume (V) is called volume strain. Volume strain = \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\)

Hooke’s Law: For small deformations, the stress is proportional to strain.

Stress ∝ strain ⇒ stress/strain = constant.

This proportional constant is called the modulus of elasticity.

Elastic Constant: The ratio of stress to strain is called the “elastic constant”. Unit: Newton/m² • D.F: ML^-1T^-2

Elastic constants are three types.

Young’s Modulus (Y): The ratio of tensile stress (or) compressive stress to longitudinal strain or compressive strain is called Young’s modulus.

Y = \(\frac{\sigma}{\varepsilon}=\frac{\text { Tensile or compressive stress }(\sigma)}{\text { Tensile or compressive strain }(\varepsilon)}\)

Shear Modulus (G): The ratio of shearing stress to the corresponding shearing strain is called shear modulus.

Shear modulus (G) = \(\frac{\text { Shear stress }\left(\sigma_{\mathrm{s}}^{\prime}\right)}{\text { Shear strain }(\theta)}\)

Bulk Modulus(B): The ratio of hydraulic stress to the corresponding hydraulic strain is called Bulk modulus.

= –\(\frac{\text { Hydraulic pressure }(F / A)}{\text { Hydraulic strain }(\Delta V / V)}\)

Compressibility (K): The reciprocal of bulk modulus is called compressibility. Compressibility K = 1/B

Poisson’s Ratio: In a stretched wire the ratio of lateral contraction strain to longitudinal elongation strain is called Poisson’s ratio.

Poisson’s ration = \(\sigma=\frac{\Delta d / d}{\Delta L \cdot / L}\)

Poisson’s ratio Is a ratio of two strains so it has only numbers.

For steel Poisson’s ratio is 0.28 to 0.30, for aluminum alloys it is up to 0.33.

Elastic Potential Energy (u): When a wire is under tensile stress, work is done against the inter-atomic forces. This work is stored in the wire in the form of elastic potential energy.

(or)

Work done to stretch a wire against inter-atomic forces is termed as “elastic potential energy”.

Elastic potential energy \((\mathrm{u})=\frac{1}{2} \frac{\mathrm{YAl} l^2}{\mathrm{~L}}=\frac{1}{2} \sigma \varepsilon\)

or \(\mathrm{u}=\frac{1}{2}\) stress x strain x volume of wire.

Ductile Materials: If the stress difference between ultimate tensile strength and fracture point is high then it is called ductile material. Example: Silver, Gold.

Brittle Material: If the stress difference between ultimate tensile strength and fracture point is very less then that substance is called brittle material. Example: Cast iron.

Elastomers: Substances that can be stretched to cause large, strains are called elastomers. Example: Rubber, Tissues of aorta.

Mechanical Properties Of Solids Solutions KSEEB Class 11 Physics Important Formulae

Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{F}}{\mathrm{A}}\); Unit: \(\mathrm{N} / \mathrm{m}^2\) (or) Pascal.

Strain \(=\frac{\text { elongation }}{\text { original length }}=\frac{\Delta \mathrm{L}}{\mathrm{L}}\); No units.

Hooke’s Law: Within elastic limit, stress \(\propto\) strain (or) \(\frac{\text { stress }}{\text { strain }}=\) constant (Elastic constant).

Young’s Modulus \((\mathrm{Y})=\frac{\text { longitudinal stress }}{\text { longitudinal strain }}\)

= \(\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{e} / \mathrm{L}}=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{e}}\).

In Searle’s apparatus \(\mathrm{Y}:=\frac{\mathrm{gL}}{\pi \mathrm{r}^2} \cdot \frac{\mathrm{M}}{\mathrm{e}}\)

1. If two wires are Hindu with same materials have lengths l₁, l₂ and radii r₁, r₂ then ratio of elongations \(\frac{e_1}{e_2}=\frac{l_1}{l_2} \cdot \frac{r_2^2}{r_1^2}\) (because \(e \propto l / r\))
2. If two wires are made with the same material and the same volume and have areas A₁ and A₂ are, subject to the same force then the ratio of elongations \(\frac{\mathrm{e}_1}{\mathrm{e}_2}=\mathrm{r}_2^4 / \mathrm{r}_1^4\) (because \(\mathrm{e} \propto \frac{1}{\mathrm{r}^4}\))
3. If two wires of the same length and area of the cross-section are subjected to the same force then the ratio of elongations \(e_1 / e_2=\frac{y_2}{y_1}\) (because \(e \propto \frac{1}{y}\))
4. If two wires are made with same material have lengths l₁ and l₂ and masses m₁ and m₂ are subjected to same force then ratio \(\frac{\mathrm{e}_1}{\mathrm{e}_2}=\frac{l_1^2}{l_2^2} \times \frac{\mathrm{m}_2}{\mathrm{~m}_1}\) (because \(\mathrm{e} \propto \frac{l^2}{\mathrm{~m}}\))

Rigidity Modulus, \(\mathrm{G}=\frac{\text { shear stress }}{\text { shear strain }}\)

= \(\frac{\mathrm{F}}{\mathrm{A} \theta}=\frac{\mathrm{FL}}{\mathrm{Al}}\)

Shear Strain \(\theta=\frac{\text { relative displacement of upper layer }}{\text { perpendicular distance between layers}}=\frac{\Delta \mathrm{x}}{\mathrm{z}}=\frac{l}{\mathrm{~L}}\)

Bulk Modulus \((B)=\frac{\text { volumetric stress }}{\text { volumetric strain }}\)

= \(\frac{F / A}{\Delta V / V}=\frac{P V}{\Delta V}\)

∴ \(\frac{1}{B}\) is called the “coefficient of compressibility” (K).

Poisson’s Ratio, \(\sigma=\frac{\text { lateral contraction strain }}{\text { longitudinal elongation }}\)

= \(-\frac{\Delta \mathrm{D} / \mathrm{D}}{\mathrm{e} / \mathrm{L}}=-\frac{\mathrm{L} \Delta \mathrm{D}}{\mathrm{D} \cdot \mathrm{e}}\)

Theoretical limits of Poisson’s ratio \(\sigma\) is -1 to 0.5; Practical limits of Poisson’s ratio \(\sigma\) is 0 to 0.5

Relations between Y, G, B, and σ are

1. \(B=\frac{Y}{3(1-2 \sigma)}\)
2. \(ŋ=\frac{Y}{2(1+\sigma)}\)
3. \(\sigma=\frac{3B-2G}{2(3B+G}\)
4. Y = \(\frac{9G B}{3 B+G}\)

The relation between volume stress and linear stress is \(\frac{\Delta V}{V}=\frac{\Delta L}{L}(1-2 \sigma)\)

Strain energy \(=\frac{1}{2} \times\) load x extension

= \(\frac{1}{2} \times F \times e\)

Strain energy per unit volume \(=\frac{1}{2} \times\) stress x strain or \(\frac{\text { stress }^2}{2 Y}\) or \(\frac{\left(\text { strain }^2\right) Y}{2}\)

When a body is heated and expansion is prevented then thermal stress will develop in the body.

Thermal stress = Y ∝ Δt

Thermal force = YA ∝ Δt

When a wire of natural length L is elongated by tensions say T₁ and T₂ have final lengths l₁ and l₂ then the Natural length of wire = L

= \(\frac{l_2 \mathrm{~T}_1-l_1 \mathrm{~T}_2}{\left(\mathrm{~T}_1-\mathrm{T}_2\right)}\)

When a wire is stretched by a load has an elongation ‘e’. If the load is completely immersed in water then decrease in elongation e¹ = Vρ gl/(AY)

where V is the volume of load, ρ = density of water.

KSEEB Class 11 Physics Chapter 10 Mechanical Properties Of Solids Very Short Answer Questions

Question 1. State Hooke’s law of elasticity.
Answer:

Hooke’s law states that within an elastic limit stress is proportional to strain.

Stress \(\propto \text { strain } \Rightarrow \frac{\text { Stress }}{\text { Strain }}=\text { Constant. }\)

This constant is known as the elastic modulus of the body.

KSEEB Class 11 Physics Mechanical Properties Of Solids Key Concepts

Question 2. State the units and dimensions of stress.
Answer:

Stress = \(\frac{\text { Porce }}{\text { Area }}=\frac{\mathrm{N}}{\mathrm{m}^2}\);

S.I Unit = \(\mathrm{Nm}^{-2}\) or pascal.

Dimensional formula \(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

Question 3. State the units and dimensions of the modulus of elasticity
Answer:

Modulus of elasticity = \(\frac{\text { Stress }}{\text { Strain }}\)

S.I Unit is \(\mathrm{Nm}^{-2}\) (or) pascal

Dimensional formula \(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

Question 4. State the units and dimensions of Young’s modulus.
Answer:

Young’s modulus, \(\mathrm{Y}=\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}\)

= \(\frac{\mathrm{F}}{\mathrm{A}} \cdot \frac{l}{\Delta l}\); SI unit = \(\mathrm{Nm}^{-2}\) (or) pascal

Dimensional formula \(\mathrm{ML}^{-1} \mathrm{~T}^2\)

Question 5. State the units and dimensions of the modulus of rigidity.
Answer:

Modulus of rigidity,

G = \(\frac{\text { shear stress }}{\text { shear strain }}=\frac{F}{A} \frac{x}{\Delta x}, \text { Sl unit }=\mathrm{Nm}^{-2}\) (or) pascal

Dimensional formula ML^-1T^-2.

Question 6. State the units and dimensions of Bulk modulus.
Answer:

Bulk modulus, \(\mathrm{B}=\frac{\text { Bulk stress }}{\text { Bulk strain }}=\frac{\Delta \mathrm{P} \cdot \mathrm{V}}{\Delta \mathrm{V}}\); SI: unit is Nm^-2 (or) pascal

Dimensional formula ML^-1T^-2.

Question 7. State the examples of nearly perfect elastic and plastic bodies.
Answer:

There is no perfectly elastic body. However, the behavior of Quartz fiber is very near to a perfectly elastic body.

Real bodies are not perfectly plastic, but the behavior of wet clay, butter, etc., can be taken as examples of perfectly plastic bodies.

Solutions For Mechanical Properties Of Solids KSEEB Physics Short Answer Questions

Question 1. Define Hooke’s law of elasticity, proportionality, permanent set, and breaking stress.
Answer:

Hooke’s Law: It states that within the elastic limit, stress is proportional to strain.

i.e., \(\text { stress } \propto \text { strain } \Rightarrow \frac{\text { stress }}{\text { strain }}=\text { constant (E). }\)

This constant is called the elastic constant (E).

Proportionality Limit: When the load is increased the elongation of the wire will also increase. The maximum load up to which the elongation is directly proportional to the load is called the proportionality limit (A). The graph is drawn between load and extension. It is a straight line ‘OA’.

Permanent Set: If the load on the wire is increased beyond the elastic limit, the elongation is not proportional to the load. On removal of the load, the wire cannot regain its original length. The length of the wire increases permanently. In the figure permanent set is given by OP. This is called a permanent set.

Breaking Stress: If the load is increased beyond the yield point the elongation is very rapid, even for small changes in load and the wire becomes thinner and breaks. This is shown as E. The breaking force per unit area is called breaking stress.

Mechanical Properties Of Solids Chapter 10 KSEEB Physics

Question 2. Define the modulus of elasticity, stress, strain, and Poisson’s ratio.
Answer:

Stress: Restoring force acting on a unit area is called stress.

∴ Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{F}}{\mathrm{A}} \text {; }\)

Unit: N/m² (or) pascal;

Dimensional formula: ML^-1T^-2.

Strain: The change in dimension per unit original dimension of a body is called strain.

∴ Strain = \(\frac{\text { elongation }}{\text { original length }}=\frac{\mathrm{e}}{l}\)

It is a ratio, so no units and dimensional formulas.

Modulus Of Elasticity: From Hooke’s Law Stress ∝Strain or \(\frac{\text { Stress }}{\text { Strain }}\) = Constant (E)

The ratio of stress to strain is called the modulus of elasticity

Unit: N/m².

Dimensional formula: ML^-1T^-2.

Poisson’s Ratio: It is defined as the ratio of lateral contraction strain to longitudinal, elongation strain.

Poisson’s ratio \((\sigma)=\frac{\text { Lateral contraction strain }}{\text { Longitudinal elongation strain }}\)

It is a ratio, so no units and dimensional.

Question 3. Define Young’s modulus, Bulk modulus, and Shear modulus.
Answer:

Young’s Modulus Y: Within the elastic limit, the ratio of longitudinal stress to longitudinal strain is called Young’s modulus

i.e., \(\mathrm{Y} =\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)

= \(\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}\)

Bulk Modulus (B): Within the elastic limit, the ratio of volumetric stress to volumetric strain is called Bulk modulus

i.e., \(B=\frac{\text { Volumetric stress }}{\text { Volumetric strain }}\)

= \(\frac{\dot{F} / \mathrm{A}}{\Delta \mathrm{V} / \mathrm{V}}=\frac{\mathrm{FV}}{\mathrm{A} \cdot \Delta \mathrm{V}}\)

Shear Modulus Or Rigidity Modulus (G): With in elastic limit, the ratio of tangential or shearing stress to shearing strain is called Shear modulus or Rigidity modulus.

i.e., \(\mathrm{G}=\frac{\text { Shearing stress }}{\text { Shearing strain }}=\frac{F / A}{\theta}=\frac{F}{A \theta}\)

Question 4. Define stress and explain the types of stress.
Answer:

Stress: When a body is subjected to a deforming force, then restoring forces will develop inside the body. These restoring forces will oppose any sort of change in its original shape. The restoring force per unit area of the surface is called stress.

Stress: Stress is defined as force applied per unit area.

Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{A}\)

D.F. = ML^-1T^-2; Unit: N/m² (or) Pascal.

Types Of Stress: It is of three types. They are:

1. Longitudinal stress
2. Tangential stress (or) Shear stress
3. Volumetric stress.

Longitudinal Stress: If the force applied on a body is along its lengthwise direction then it is called longitudinal stress. It produces deformation in length.

Longitudinal stress = \(\frac{\text { Force }}{\text { Area }}\)

Tangential Stress (Or) Shear Stress: Force applied per unit area parallel to the surface of a body trying to displace the upper layers of the body is called shearing stress.

Shear (or) tangential stress = \(\frac{\text { Force }}{\text { Area }}\) (Parallel to the surface layers)

Volumetric Stress: If force is applied on all the sides of a body or on the volume of a body then it is called volumetric stress.

Volumetric stress = \(\frac{\text { Force }}{\text { Area }}\) = (applied on volume)

Question 5. Define strain and explain the types of strain.
Answer:

Strain: Strain is defined as deformation produced per unit dimension. It is a ratio. So no units.

Strain (e) = \(\frac{\text { Elongation }}{\text { Original length }}=\frac{\Delta \mathrm{L}}{\mathrm{L}} \text {. }\)

Types Of Strain: Strain is of three types. They are:

1. Longitudinal strain
2. Tangential strain or Shear strain
3. Volumetric strain

Longitudinal Strain: The ratio of elongation to original length along a lengthwise direction is defined as longitudinal strain.

Longitudinal strain \(\mathrm{e}=\frac{\text { elongation in length }}{\text { original length }}\) = \(\frac{\Delta l}{l}\)

Bulk strain = Change in volume per unit

volume = \(\frac{\text { Change in volume }}{\text { Original volume }}=\frac{\Delta \mathrm{V}}{\mathrm{V}}\)

Shearing Strain: If the force applied on a body produces a change in shape only it is called shearing force. The angle through which a plane originally perpendicular to the fixed surface shifts due to the application of shearing stress is called shearing strain or simplified shear θ.

Shear strain = \(\theta=\left(\frac{\Delta l}{l}\right)=\tan \theta\) (since is a very small tan θ = θ)

Question 6. Define strain energy and derive the equation for the same.
Answer:

Strain Energy: The energy developed in (string), a body when it is strained is called strain energy.

Let a force F be applied on the lower end of the wire, fixed at the upper end. Let the extension be dl.

∴ Work done \(=\mathrm{dW}=\mathrm{Fd}\)

Total work done in stretching it from 0 to l = \(W=\int_0^1 \mathrm{dW}=\int_0^1 \mathrm{Fd} l\)

∴ W = \(\int_0^1 \frac{Y a l}{L} d l=\frac{Y a l^2}{2 L}=\frac{1}{2} \frac{\mathrm{Yal}}{L} \cdot l\)

(because \(\mathrm{F}=\frac{\mathrm{Yal}}{\mathrm{L}}\))

W = \(\frac{1}{2} \times \text { stretching force } \times \text { elongation }\)

= \(\frac{1}{2} \times \frac{\text { stretching force } \times \text { elongation }}{\mathrm{aL}}\)

= \(\frac{1}{2} \times \text { stress } \times \text { strain }\)

Mechanical Properties Of Solids Questions And Answers KSEEB Physics

Question 7. Explain why steel is preferred to copper, brass, and aluminum in heavy-duty machines and in structural designs.
Answer:

For metals Young’s moduli are large. Therefore, these materials require a large force to produce a small change in length. To increase the length of a thin steel wire of 0.1 cm² cross-sectional area by 0.1 %, a force of 2000 N is required.

The force required to produce the same strain in aluminum, brass, and copper wire having the same cross-sectional area are 690 N, 900N, and 1100 N respectively.

It means that steel is more elastic than copper, brass, and aluminum. It is for this reason that steel is preferred, in heavy-duty machines and in structural designs.

Question 8. Describe the behavior of a wire under a gradually increasing load.
Answer:

Behavior Of A Wire Under Increasing Load: Let a wire be suspended at one end and loads are attached to the other end. When loads are attached to the other end. When loads are gradually increased the following changes are noticed.

1. Proportionality Limit (A): When the load has increased the elongation of the wire gradually increases. The maximum load up to which the elongation is directly proportional to the load is called the proportionality limit (A). In this region, Hooke’s I Law is obeyed.
2. Elastic Limit (B): If the load is Increased above the proportionality limit the elongation is not proportional to the load. Hooke’s law is not obeyed. The maximum load on the wire up to which it exhibits elasticity is called the elastic limit (B in the graph).
3. Permanent Set (C): If the load on the wire is increased beyond the elastic limit say up to ‘C’, the elongation is not proportional to the load. On removal of the load, the wire does not regain its original length. The length of the wire increases permanently. In the figure permanent set is given by OP. So OP is called a permanent set.
4. Point Of Ultimate Tensile Strength (D): If the load is further increased, up to ‘D’ then strain increases rapidly even though there is no increase in stress. Elongation without an increase in load is called creeping. This behavior of metal is called yielding.
5. Fracture Point (E): If the load is increased beyond the Yield point the elongation is very rapid, even for small changes in load the wire becomes thinner and breaks. This is shown as E. The breaking force per unit area is called breaking stress.

Question 9. Two identical solid balls, one of ivory and the other of wet-clay Eire dropped from the same height onto the floor. Which one will rise to a greater height after striking the floor and why?
Answer:

We know that ivory bail is more elastic than wet-clay ball. Therefore, the ivory ball will tend to regain its original shape in a very short time after the collision. Due to this, there will be a large energy and momentum transfer to the ivory ball in comparison to the wet-clay ball. As a result of it, the ivory ball will rise higher after the collision.

Question 10. While constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends. Why?
Answer:

The use of pillars or columns is very common in buildings and bridges. A pillar with rounded ends supports less load than that with a distributed shape at the ends. Hence, for this reason, while constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends.

Question 11. Explain why the maximum height of a mountain on Earth is approximately 10 km.
Answer:

A mountain base is not under uniform compression and this provides some shearing stress to the rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.

At the bottom of a mountain of height h’, the force per unit area due to the weight of the mountain is hρg where p is the density of the material of the mountain and g is the acceleration due to gravity.

The material at the bottom experiences this force in the vertical direction and the sides of the mountain are free. There is a shear component, approximately hρg itself. Now the elastic limit for a typical rock is 3 x 10⁷ Nm^-2. Equating this to hρg with ρ = 3 x 10³ kg m^-3 gives

h = \(\frac{30 \times 10^7}{3 \times 10^3 \times 10}=10 \mathrm{~km}\)

Hence, the maximum height of a mountain on Earth is approximately 10 km.

Question 12. Explain the concept of Elastic Potential Energy in a stretched wire and hence obtain the expression for it.
Answer:

When a wire is put under tensile stress, work is done against the inter-atomic forces. This work is stored in the form of Elastic potential energy.”

Expression To Elastic Potential Energy: To stretch a wire, force is applied. As a result, it elongates. So the force applied is useful to do some work. This work is stored in it as potential energy. When the deforming force is removed, this energy is liberated as heat. The energy developed in a body (string) when it is strained is called strain energy.

Let a force F be applied on a wire fixed at the upper end. Let the extension be dl.

∴  Work done = dW = Fdl

Total work done in stretching from 0 to l = \(W=\int_0^l \mathrm{~d} W=\int_0^l \mathrm{Fd} l\)

∴ W = \(\int_0^l \frac{\mathrm{Ya} l}{\mathrm{~L}} \mathrm{~d} l=\frac{\mathrm{Ya} l^2}{2 \mathrm{~L}}=\frac{1}{2} \frac{\mathrm{Ya} l}{\mathrm{~L}} \cdot l\)

(because \(\mathrm{F}=\frac{\mathrm{Ya} l}{\mathrm{~L}}\))

W = \(\frac{1}{2} \times\) stretching force x elongation

∴ Work done per unit volume

= \(\frac{1}{2} \times \frac{\text { stretching force } \times \text { elongation }}{A \times L}\)

= \(\frac{1}{2} \times \text { stress } \times \text { strain }\)

KSEEB Class 11 Physics Chapter 10 Mechanical Properties Of Solids

KSEEB Physics Class 11 Mechanical Properties of Solids Long Answer Questions

Question 1. Define Hooke’s law of elasticity and describe an experiment to determine Young’s modulus of the material of a wire.
Answer:

Hooke’s Law: Within elastic limit, stress is directly proportional to strain.

∴ strain \(\propto\) stress

∴ \(\frac{\text { stress }}{\text { strain }}=\text { constant }(E)\)

where E constant is called modulus or elasticity of the material of n body.

Determination Of Young’s Modulus Of A Wire: The apparatus used to find Young’s modulus of a wire consists of two long wires A and B of the same length made with the same material. These two wires are suspended from a rigid support and a vernier scale Y is attached to them. Wire A is connected to the main scale (M).

A fixed load is connected to this cord to keep tension in the wire. This is called reference wire. The second wire ‘B’ is connected to the vernier scale ‘V’. An adjustable load hanger is connected to this wire. This is called experimental wire.

Procedure: Let a load M₁ be attached to the weight hanger at the vernier. Main Scale Reading (M.S.R) and Vernier Scale Reading (V.S.R) are noted. Weights are gradually increased in the steps of \(\frac{1}{2}\) kg up to a maximum load of say 3 kg. Every time M.S.R and V.S.R are noted. They are placed in tabular form.

Now loads are gradually decreased in steps of \(\frac{1}{2}\) kg. While decreasing M.S.R and V.S.R are noted for every load, values are posted in tabular form.

Let 1st reading with mass M₁ is e₁ and 2nd reading with mass M₂ is e₂.

Change in load M = M₂ – M₁

elongation e = e₂-e₁

‘M’ and ‘e’ values are calculated and a graph is plotted.

Average m/e

Force on the wire = mg

Area of cross unction of the wire = πr² (r = radius of the wire)

Elongation = e

Original length = l

Stress = \(\left(\frac{\mathrm{F}}{\mathrm{a}}\right)=\frac{\mathrm{mg}}{\pi \mathrm{r}^2}\)

Strain \(=\left(\frac{\mathrm{e}}{l}\right)\)

Young’s modulus, \(\mathrm{Y}=\left(\frac{\text { stress }}{\text { strain }}\right)\)

Y = \(\left(\frac{\mathrm{mg} l}{\mathrm{e} \pi \mathrm{r}^2}\right)=\left(\frac{\mathrm{g} l}{\pi \mathrm{r}^2}\right)\left(\frac{\mathrm{m}}{\mathrm{e}}\right)\)

A graph between load (m) and elongation ‘e’, is a straight line passing through the origin.

The slope of the graph (tan θ) gives \(\frac{m}{e}\). This value is substituted in the above equation to find Young’s modulus of the material of the wire.

Precautions:

1. The load applied should be much smaller than the elastic limit.
2. Reading is noted only after the air bubble is brought to the center of spirit level.

Chapter 10 Mechanical Properties Of Solids KSEEB Problems

Question 1. A copper wire of 1mm diameter is stret¬ched by applying a force of 10 N. Find the stress in the wire.
Solution:

Diameter, d = 1mm

∴ radius, r = 0.5 mm = 0.5 x 10^-3 m

Force, F = 10N

Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{\pi r^2}\)

∴ Stress = \(\frac{10}{3.141 \times 0.5 \times 0.5 \times 10^{-6}}\)

= \(\frac{40 \times 10^6}{3.141}=1.273 \times 10^7 \mathrm{~Pa}\)

Question 2. A tungsten wire of length 20cm is stretched by 0.1cm. Find the strain on the wire.
Solution:

Length of wire, l =20cm =0.2m elongation, e = 0.1cm = 1 x 10^-3 m

Strain = \(\frac{\text { elongation }}{\text { original length }}=\frac{1 \times 10^{-3}}{0.2}\)

= \(5 \times 10^{-3}=0.005\)

Question 3. If an iron wire is stretched by 1 %, what is the strain on the wire?
Solution:

Increase in length = e = 1% = \(\frac{1}{100}\) l

∴ Strain = \(\frac{\text { elongation }}{\text { original length }}=\frac{\mathrm{e}}{l}\)

= \(\frac{1 \times l}{100 \times l}=\frac{1}{100}=0.01\)

Question 4. A brass wire of a cross-sectional area of 2mm is suspended from a rigid support and a body of volume 100cm³ is attached to its other end. If the decrease in the length of the wire is 0.11mm, when the body is completely immersed in water, find the natural length of the wire.
Solution:

Length of wire, l = 2m

Diameter, d = 1 mm = 10 m^-3

Force, F = 20N

Increase in length, e = 0.51 mm = 0.51 x 10^-3 m

Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{\pi \mathrm{d}^2 / 4}=\frac{20 \times 4}{3.141 \times 10^{-6}}\)

= \(2.546 \times 10^7 \mathrm{~N} / \mathrm{m}^2\)

Strain = \(\frac{\text { Elongation }}{\text { Length }}=\frac{\mathrm{e}}{\mathrm{L}}=\frac{0.51 \times 10^{-3}}{2}\)

= \(0.255 \times 10^{-3} \mathrm{~m}\)

Young’s modulus, Y = \(\frac{\text { stress }}{\text { strain }}\)

= \(\frac{2.546 \times 10^7}{0.255 \times 10^{-3}}=9.984 \times 10^{10} \mathrm{~N} / \mathrm{m}^2\)

Question 5. A copper wire and an aluminum wire have lengths in the ratio 3: 2, diameters in the ratio 2 : 3, and forces applied in the ratio 4: 5. Find the ratio of increase in length) of the two wires. (Y_Cu = 1.1 x 10¹¹ Nm^-2, Y_Al = 0.7 x 10¹¹ Nm^-2)
Solution:

Ratio of lengths, l₁: l₂ = 3: 2 ;

The ratio of diameters, d₁: d₂ = 2 : 3

Ratio of forces, F₁: F₂ = 4 : 5

Y₁ = Y of copper = 1.1 x 10¹¹

Y₂ = Y of Aluminium = 0.7 x 10¹¹;

The ratio of elongation, e₁: e₂?

e = \(\frac{\mathrm{Fl}}{\mathrm{AY}}=\frac{4 \mathrm{~F} l}{\pi \mathrm{d}^2 \mathrm{Y}}\)

∴ \(\frac{\mathrm{e}_1}{\mathrm{e}_2}=\frac{4 \mathrm{~F}_1 l_1}{\pi \mathrm{d}_1^2 \mathrm{Y}_1} \times \frac{\pi \mathrm{d}_2^2 \mathrm{Y}_2}{4 \mathrm{~F}_2 l_2}\)

⇒ \(\frac{\mathrm{e}_1}{\mathrm{e}_2}=\frac{\mathrm{F}_1 l_1 \mathrm{~d}_2^2 \mathrm{Y}_2}{\mathrm{~F}_2 l_2 \mathrm{~d}_1^2 \mathrm{Y}_1}=\frac{4 \times 3 \times 3^2 \times 0.7 \times 10^{11}}{5 \times 2 \times 2^2 \times 1.1 \times 10^{11}}\)

= \(\frac{189}{110}\)

∴ \(\mathrm{e}_1: \mathrm{e}_2=189: 110\)

Question 6. A brass wire of cross-sectional area 2mm² is suspended from a rigid support and a body of volume 100cm³ is attached to its other end. If the decrease in the length of the wire is 0.11mm, when the body is completely immersed in water, find the natural length of the wire. (Y_brass =0.91 x 10¹¹ Nm^-2 ρ_water = 10³ kg m^-3)
Solution:

Area of cross-section, A = 2mm² = 2 x 10^-6 m²

Volume of body, V = 100 cc = 100 x 10^-6 m³

Decrease in length, e’ = 0.11mm = 0.11 x 10^-3 m

Young’s modulus of brass, Y = 0.91 x 10¹¹ N/m²

Density of water, \(\rho=1000 \mathrm{~kg} / \mathrm{m}^3\)

Use \(\mathrm{e}^{\prime}=\frac{\mathrm{V} \rho g l}{\mathrm{AY}}\)

Natural length of wire, l = \(\frac{e^{\prime} \mathrm{AY}}{V_{\rho g}}\)

= \(\frac{0.11 \times 10^{-3} \times 2 \times 10^{-6} \times 0.91 \times 10^{11}}{100 \times 10^{-6} \times 1000 \times 9.8}\)

∴ I = \(\frac{0.2002 \times 10^2}{9.8}=\frac{20.02}{9.8}=2.043 \mathrm{~m}\)

Question 7. There are two wires of the same material. Their radii and lengths are both in the ratio 1: 2. If the extensions produced are equal, what is the ratio of the loads?
Solution:

Ratio of lengths, l₁ : l₂ = 1: 2

The ratio of radii, r₁: r₂ = 1: 2

Extensions produced are equal ⇒ e₁= e₂;

Made of same material ⇒ Y₁ = Y₂

Ratio of loads m₁: m₂ =? mg

Use \(\mathrm{Y}=\frac{\mathrm{mg}}{\pi \mathrm{r}^2} \frac{l}{\mathrm{e}}\) then \(\frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\mathrm{Y}_1 \mathrm{r}_1^2 l_2}{\mathrm{l}_1 \mathrm{Y}_2 \mathrm{r}_2^2}=\frac{1^2 \times 2}{1 \times 2^2}=\frac{1}{2}\)

∴ \(\mathrm{m}_1: \mathrm{m}_2=1: 2\)

Question 8. Two wires of different materials have the same lengths and areas of cross-section. What is the ratio of their increase in length when the forces applied are the same? (Y₁ = 0.90 x 10¹¹ Nm^-2, Y₂ = 3.60 x 10¹¹ Nm^-2)
Solution:

Lengths are same ⇒ l₁= l₂;

The area of cross sections are same, ⇒ A₁ = A₂

Y₁ = 0.9 x 10¹¹ N/m²

Y₂ = 3.60 x 10¹¹ N/m²

Elongation, e = \(\frac{F l}{A Y}\)

∴ \(\frac{e_1}{e_2}=\frac{F_1 l_1}{A_1 Y_1} \cdot \frac{A_2 Y_2}{F_2 l_2}\)

(because F, l, and A are the same)

∴ \(\frac{e_1}{e_2}=\frac{Y_2}{Y_1}\)

∴ \(\frac{e_1}{e_2}=\frac{3.60 \times 10^{11}}{0.9 \times 10^{11}}=\frac{4}{1} \text { or } \mathrm{e}_1: \mathrm{e}_2=4: 1\)

Question 9. A metal wire of length 2.5in and area of cross-section 1.5 x 10^-6 m² is stretched through 2mm. If Young’s modulus is 1.25 x 10¹¹ Nm^-2, find the tension In the wire.
Solution:

Length of wire, l = 2.5m

Y = 1.25 x 10¹¹ N/m²

Area of cross-section, A = 1.5 x 10^-6 m²

Elongation, e = 2 m.m = 2 x 10^-3 m

Tension, T = mg = F =?

Y= \(\frac{\mathrm{Fl}}{\mathrm{Ae}} \Rightarrow \mathrm{F}=\frac{\mathrm{YAe}}{l}\)

∴ Tension, \(\mathrm{T}=\frac{1.25 \times 10^{11} \times 1.5 \times 10^{-6} \times 2 \times 10^{-3}}{2.5}\)

= \(\frac{3.75 \times 10^2}{2.5}=150 \mathrm{~N}\)

Question 10. An aluminum wire and a steel wire of the same length and cross-section are joined end-to-end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35 mm, find the ratio of the

1. Stress in the two wires and
2. Strain in the two wires.

(Y_Al=0.7 x10¹¹ Nm^-2, Y_steel = 2 x 10¹¹ Nm^-2)

Solution:

1. Length is same ⇒ l₁ = I₂

Area is same ⇒ A₁ = A₂

In composite wire, the same load will act on both wires.

∴ Ratio of stress = 1:1

2. Total elongation, e = 1.35mm = e_Al+ e_s

Young’s modulus of aluminium = 7 x 10¹⁰ N/m²,

Y of steel = 2 x 10¹¹ N/m²

Elongation, e = \(\frac{F l}{A Y}\)

But F, l, and A are the same

∴ \(\mathrm{e} \propto \frac{1}{\mathrm{Y}} \text { or } \frac{\mathrm{e}_{\mathrm{Al}}}{\mathrm{e}_{\mathrm{s}}}=\frac{\mathrm{Y}_{\mathrm{s}}}{\mathrm{Y}_N}=\frac{20 \times 10^{10}}{7 \times 10^{10}}=\frac{20}{7}\)

∴ The ratio of strains in the wires is 20: 7

Question 11. A 2 cm cube of some substance has its upper face displaced by 0.15cm due to a tangential force of 0.3 N while keeping the lower face fixed. Calculate the rigidity modulus of the substance.
Solution:

Side of cube, a = 2.0 cm = 2 x10^-2 m

∴ Area, A = 4 x 10^-4 m²

Displacement of upper layer = 0.15cm = 0.15 x 10^-2 m

Tangential force, F = 0.30N

Rigidity modulus, \(\eta=\frac{\mathrm{F}}{\mathrm{A}} \cdot \frac{\mathrm{x}}{\Delta \mathrm{x}}\)

= \(\frac{0.30}{4 \times 10^{-4}} \frac{2 \times 10^{-2}}{0.15 \times 10^{-2}}\)

∴ \(\eta=\frac{0.60 \times 10^4}{0.60}=1 \times 10^4 \mathrm{~N} / \mathrm{m}^2\)

Question 12. A spherical ball of volume 1000 cm³ is subjected to a pressure of 10 atmospheres. The change in volume is 10^-8 cm³. If the ball is made of iron, find its bulk modulus. (1 atmosphere = 1 x 10⁵ Nm^-2)
Solution:

Volume of ball, V = 1000 cm³ = 10^-3 m³ (1M³ = 10⁶ cm³)

Pressure, P = 10 atmospheres = 10 x 10⁵ pa (1 atm = 10⁵ pascal)

Change in volume, ΔV = 10^-8 cm³

K = \(\frac{\mathrm{PV}}{\Delta \mathrm{V}}=\frac{10^6 \times 10^{-3}}{10^{-8}}=1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)

Question 13. A copper cube of the side of length 1 cm is subjected to a pressure of 100 atmospheres. Find the change in its volume if the bulk modulus of copper is 1.4 x 10¹¹ Nm^-2 (1 atm = 1 x 10⁵ Nm^-2).
Solution:

Side of cube,‘a’ = 1 cm = 10^-2 m

∴ Volume of cube = 10^-6 m

Pressure, P = 100 atm = 100 x 10⁵ = 10⁷ pa

Bulk modulus, K = 1.4 x 10¹¹ N/m²;

Change in volume, \(\Delta V=\frac{P. V}{K}\)

= \(\frac{10^7 \times 10^{-6}}{1.4 \times 10^{11}}=\frac{10^{-10}}{1.4}=0.7143 \times 10^{-10} \mathrm{~m}^3\)

KSEEB Class 11 Physics solutions for Chapter 10 Mechanical Properties of Solids

Question 14. Determine the pressure required to reduce the given volume of water by 2%. The bulk modulus of water is 2.2 x 10⁹ Nm^-2
Solution:

Bulk strain = \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=2 \% \Rightarrow \Delta \mathrm{V}=\frac{2}{100} \mathrm{~V}\)

Bulk modulus, \(\mathrm{K}=2.2 \times 10^9 \mathrm{~N} / \mathrm{m}^2\)

∴ Pressure required, \(\mathrm{P}=\frac{\mathrm{K} \Delta \mathrm{V}}{\mathrm{V}}\)

= \(\frac{2.2 \times 10^9 \times 2 \mathrm{~V}}{\mathrm{~V} \times 100}=4.4 \times 10^7 \text { pascal }\)

Question 15. A steel wire of length 20 cm is stretched to increase its length by 0.2 cm. Find the lateral strain in the wire if the Poisson’s ratio for steel is 0.19.
Solution:

Length of wire, l = 20, cm = 0.20m,

Poisson’s ratio, σ =0.19

Increase in length, Δl = 0.2 cm = 2 x 10^-3 m

lateral strain = ?

Lateral strain = σ x longitudinal strain ‘e’

But e = \(\frac{\Delta l}{l}\)

∴ Lateral strain = \(\sigma \frac{\Delta l}{l}=\frac{0.19 \times 2 \times 10^{-3}}{0.20}\)

= \(1.9 \times 10^{-3}=0.0019 \mathrm{~m}\).

KSEEB Solutions For Class 11 Physics Chapter 8 Gravitation Important Points

Kepler’s Laws:

Law Of Orbits (1st Law): All planets move in an elliptical orbit with the sun at one of its foci.

Law Of Areas (2nd Law): The line joining the planet to the sun sweeps equal areas in equal intervals of time, i.e., \(\frac{\Delta \mathrm{A}}{\Delta \mathrm{T}}\) = constant.

i.e., planets will appear to move slowly when they are away from the sun, and they will move fast when they are nearer to the sun.

Law Of Periods (3rd Law): The square of time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

i.e., \(T^2 \propto R^3 \Rightarrow \frac{T^2}{R^3}=\text { constant }\)

Read and Learn More KSEEB Class 11 Physics Solutions

Newton’s Law Of Gravitation (Or) Universal Law Of Gravitation: Everybody in the universe attracts other bodies with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

F \(\propto m_1 m_2, F \propto \frac{1}{r^2} \Rightarrow F=G \frac{m_1 m_2}{r^2}\)

Central Force: A central force is a force which acts along the line joining the sun and the planet or along the line joining the two mass particles.

Conservative Force: For a conservative force work done is independent of the path. Work done depends only on initial and final positions. Gravitational force is a conservative force.

Gravitational Potential Energy: Potential energy arising out of gravitational force is called gravitational potential energy.

Since gravitational force is a conservative force gravitational potential depends on the position of the object.

V = –\(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}}\)

Gravitational Potential: Gravitational potential due to the gravitational force of the earth is defined as the “potential energy of a par tide of unit mass at that point”.

Gravitational potential V = \(\frac{GM}{r}\)

(r = distance from the centre of the earth)

Acceleration Due To Gravity (g): \(\frac{G N}{R^2}\)

Acceleration Due To Gravity Below And Above Surface Of Earth:

For points above the earth, the total mass of the earth seems to be concentrated at the centre of the earth.

For a height ‘h’ above the earth \(g(h)=\frac{G M_E}{\left(R_E+h\right)^2}\)

When h<<\(R^E\)

g(h) = \(g\left(1+\frac{h}{R_E}\right)^{-2} \simeq g\left(1-\frac{2 h}{R_E}\right)\)

For a point inside the earth at a depth ‘d’ below the groundmass of the earth (M_s) with radius (R_E– d) Is considered. That mass seems to be at the centre of the earth.

g’ = \(g\left(1-\frac{d}{R}\right)\)

Escape speed (v)_min: The minimum initial velocity on the surface of the earth to overcome gravitational potential energy is defined as “escape speed v_e.”

∴ \(\mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)

Orbital Velocity: The velocity of a body revolving in the orbit is called orbital velocity.

∴ Orbital velocity \(V_0=\sqrt{\frac{\mathrm{GM}}{R}} \Rightarrow V_0=\sqrt{g R}\)

1. Relation between orbital velocity V and escape speed v_e = √2 V₀
2. If the velocity of a satellite in the orbit is increased by √2 times or more it will go to infinite distance.

Time Period Of The Orbit (T): The time taken by a satellite to complete one rotation in the orbit is called “time period of rotation”.

T = \(2 \pi \frac{\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3 / 2}}{\sqrt{\mathrm{GM}_{\mathrm{E}}}}\)

Geostationary Orbit: For a geostationary orbit in the equatorial plane its time period of rotation is 24 hours, i.e., the angular velocity of the satellite in that orbit is equal to the angular velocity of the rotation of the earth. Geostationary orbit is at a height of 35800 km from Earth.

Geostationary Satellite: A Geostationary satellite will revolve above the earth in a geostationary orbit along the direction of rotation of the earth. So it always seems to be stationary w.r.t earth.

The period of geostationary satellites is 24 hours. It rotates in an equatorial plane in the west-to-east direction.

Polar Satellites: Polar satellites are low-attitude satellites with an altitude of 500 km to 800 km. They will revolve in the north-south direction of the earth. Time period of polar satellites is nearly 100 minutes.

Weightlessness: For a freely falling body its weight seems to be zero. The weight of a body falling downwards with acceleration ‘a’ is w’ = mg’ = m(g-a). When a = g the body is said to be under free fall and it seems to be weightless.

Chapter 8 Gravitation Solutions KSEEB Class 11 Physics Important Formulae

Force between two mass particles, \(\mathrm{F}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}\)

Universal gravitational constant, \(\mathrm{G}=\frac{\mathrm{Fr}^2}{\mathrm{~m}_1 \mathrm{~m}_2}\)

G = \(6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{Kg}^2\)

D.F.: \(\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\)

Relation between g and G is, \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}=\frac{4}{3} \pi \rho \mathrm{G} \cdot \mathrm{R}\)

Variation of g with depth, \(g_d=g\left(1-\frac{d}{R}\right)\)

Variation of g with height, \(g_h=g\left(1-\frac{2 h}{R}\right)\)

For small values of h i.e., h < < R then \(g_h=g\left(1-\frac{2 h}{R}\right)\)

1. Gravitational potential, U = \(-\frac{\mathrm{GMm}}{\mathrm{R}}\)
2. If a body is taken to a height ‘h’ above the ground then

Gravitational potential, \(U_h=-\frac{G M m}{(R+h)}\)

Orbital velocity, \(V_0=\sqrt{\frac{G M}{R}}=\sqrt{g R}\)

Orbital angular velocity, \(\omega_0=\sqrt{\frac{G M}{R^3}}=\sqrt{\frac{g}{R}}\)

Escape velocity, \(\mathrm{V}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gR}}\)

Time period of geostationary orbit = 24 hours.

Angular velocity of earth’s rotation \((\omega)=\frac{2 \pi}{24 \times 60 \times 60}=0.072 \times 10^{-3} \mathrm{rad} / \mathrm{sec} \text {. }\)

KSEEB Class 11 Physics Chapter 8 Gravitation Very Short Answer Questions

Question 1. State the unit and dimension of the universal gravitational constant (G).
Answer:

Units of G = N-m²/ kg².

Dimensional formula = M^-1 L³ T^-2.

Question 2. State the vector form of Newton’s law of gravitation.
Answer:

Vector form of Newton’s Law of gravitation \(\overline{\mathrm{F}}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2 \overline{\mathrm{r}}}{\overline{\mathrm{r}}^3}\)

Question 3. If the gravitational force of the Earth on the Moon is F. What is the gravitational force of the Moon on the Earth? Do these forces form an action-reaction pair?
Answer:

The gravitational force between the earth and moon and moon and earth are the same i.e., F_EM = – F_ME

The gravitational force between the bodies is treated as an action-reaction pair.

KSEEB Class 11 Physics Gravitation

Question 4. What would be the change in acceleration due to gravity (g) at the surface, if the radius of Earth decreases by 2% keeping the mass of Earth constant?
Answer:

Acceleration due to gravity, g = \(\frac{GM}{R^2}\)

When the mass is kept constant and the radius is decreased by 2% then \(\frac{\Delta R}{R} \times 100=2\)

From the distribution of errors in multiplications and divisions \(\frac{\Delta g}{g}\) x 100 = -2 \(\frac{\Delta R}{R}/100\)

% Change in g = – 2 x 2 = – 4% – ve sign indicates that when R decreases ‘g’ increases.

Question 5. As we go from one planet to another, how will

1. The mass and
2. Does the weight of the body change?

Answer:

1. As we go from one planet to another planet mass of the body does not change. The mass of a body is always constant.
2. As we move from one planet to another planet weight of the body gradually decreases. It becomes weightless. When we approach the other planet the weight will gradually increase.

Question 6. Keeping the length of a simple pendulum constant, will the time period be the same on all planets?
Answer:

Even though the length of pendulum l is the same T the time period of oscillation T value changes from planet to planet.

Time period of pendulum, \(\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}\)

i.e., T depends on l and g.

Acceleration due to gravity, \(\left(g=\frac{G M}{R^2}\right)\) changes from planet to planet.

Hence Time period of the pendulum changes even though the length ‘l’ is the same.

Question 7. Give the equation for the value of g at a depth ’d’ from the surface of Earth. What is the value of ‘g’ at the centre of Earth?
Answer:

Acceleration due to gravity at a depth d’ below the ground is, g_d = g\(\left(1-\frac{D}{R}\right)\)

Acceleration due to gravity at the centre of the earth is zero. (Since D = R)

Question 8. What are the factors that make ‘g’ the least al the equator amt maximum at the poles?
Answer:

‘g’ Is Least At Equator Due To

The equatorial radius of Earth is the maximum

∴ g = \(\frac{G M}{R^2}\) (R = maximum)

Due to the rotation of the earth, centrifugal force will act on the bodies. It opposes the gravitational pull of the earth on the bodies. At the equator centrifugal force is maximum. So the ’g’ value is least at the equator.

The ’g’ ⇒ Maximum At Poles Due To

The polar radius of the earth is minimum (because \(g=\frac{G M}{R^2}\))

Centrifugal force due to the rotation of the earth is zero at the poles. This centrifugal force reduces Earth’s gravitational pull.

Since the Centrifugal force is zero, the ‘g’ value is maximum at the poles.

KSEEB Class 11 Physics Chapter 8 Gravitation

Question 9. “Hydrogen is in abundance around the sun but not around the earth”. Explain.
Answer:

The escape velocity of the sun is very high compared to that of the Earth. The gravitational pull of the sun is very large because of its larger mass compared to that of the Earth.

So it is very difficult for hydrogen to escape from the Sun’s atmosphere. Hence hydrogen is abundant in the sun.

Question 10. What is the time period of revolution of a geostationary satellite? Does it rotate from West to East or from East to West?
Answer:

Time period of a geostationary satellite is equal to time period of rotation of the earth.

Time period of geostationary orbit T = 24 hours. Satellites in geostationary orbit will revolve around the earth in west to east direction in an equatorial plane.

Question 11. What are polar satellites?
Answer:

Polar Satellite: Polar satellites are low-altitude satellites. They will revolve around the poles of the earth in a north-south direction. Time period of polar satellites is nearly 100 minutes.

KSEEB Class 11 Physics Solutions For Chapter 8 Gravitation Short Answer Questions

Question 1. State Kepler’s Laws of planetary motion
Answer:

Kepler’s Laws:

Law Of Orbits (1st Law): All planets will move in elliptical orbits with the sun lying al one of its foci.

‘a’ and ‘b’ are lengths of semi-major axis and semi-minor axis.

Law Of Areas (2nd Law): The line that joins any planet to the sun sweeps equal areas in equal intervals of time, i.e., \(\frac{\Delta \mathrm{A}}{\Delta T}\) = constant

i.e., planets will appear to move slowly when they are away from the sun, and they will move fast when they are nearer to the sun.

Law Of Periods (3rd Law): The square of time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

i.e., \(\mathrm{T}^2 \propto \mathrm{a}^3 \text { or } \frac{\mathrm{T}^2}{\mathrm{a}^3}=\text { constant }\)

where T = time period of revolution a = semi-major axis (length)

Question 2. Derive the relation between acceleration due to gravity (g) at the surface of a planet and Gravitational constant (G).
Answer:

Relation Between g and G: Each and everybody was attracted towards the centre of the earth with some force. This is called the weight of the body, W = mg…..(1)

This force is due to the gravitational pull on the body by the Earth.

For small distances above the earth from the centre of the earth Is equal to the radius of the earth ‘R’.

According to Newton’s law of gravitation, w = mg

Force between two bodies, \(F=\frac{G M m}{R^2} \text {. }\)…..(2)

From equations, (1) and (2)

∴ \(\mathrm{mg}=\frac{\mathrm{GMm}}{\mathrm{R}^2} \Rightarrow \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\)

∴ Relation between \(\mathrm{g}\) and G is \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\)

Question 3. How does the acceleration due to gravity change for the same values of height (h) and depth(d)?
Answer:

Variation Of ‘g’ With Altitude: When we go to a height ‘h’ above the ground ‘g’ value decreases.

On surface of earth (g) = \(\frac{GM}{R^2}\)

At an altitude ‘h’ \(g(h)=\frac{G M}{(R+h)^2}\) because R + h is the distance from the centre of the earth to the given point at ‘h

h <<R ⇒ g(h) = \(g\left(1-\frac{2 h}{R}\right)\)

So acceleration due to gravity decreases with height above the ground.

Variation of ‘g’ With Depth: When we go deep into the ground ‘g’ value decreases.

At a depth ’d’ inside the groundmass of the earth up to point d from the centre will exhibit force of attraction on the body.

The remaining mass does not exhibit any influence. So effective radius is (R – d) only. Acceleration due to gravity ’g’ at a depth ’d’ is given by

∴ \(\mathrm{g}_{\mathrm{d}}=\frac{4}{3} \pi \rho \mathrm{G}(\mathrm{R}-\mathrm{d})\) (because \(\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{GR}\))

(or) \(\mathrm{g}_{\mathrm{d}}=\frac{4}{3} \pi \rho \mathrm{GR}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\)

(or) \(g_d=g\left(1-\frac{d}{R}\right)\)

So, the ‘g’ value decreases with depth below the ground.

Question 4. What is orbital velocity? Obtain an expression for it.
Answer:

Orbital Velocity (V₀): The velocity of a satellite moving in the .orbit is called orbital velocity (V₀).

Let a satellite of mass m is revolving around the earth in a circular orbit at a height ‘h’ above the ground.

The radius of the orbit = R + h where R is the radius of the earth.

In orbital motion is “The centrifugal and centripetal forces acting on the satellite”.

Centrifugal force = \(\frac{m V^2}{r}=\frac{m V_0^2}{R+h}\)…..(1)

(In this case \(V=V_0\) and r=R+h)

Centripetal force is the force acting towards the centre of the circle it is provided by the gravitational force between the planet and the satellite.

∴ F = \(\frac{G M \cdot m}{(R+h)^2}\)….(2)

(1)=(2) \(\frac{m V_0^2}{(R+h)}=\frac{G M \cdot m}{(R+h)^2}\)

∴ \(V_0^2=\frac{G M}{R+h}\) or \(V_0=\sqrt{\frac{G M}{R+h}}\)

When h<<R then orbital velocity,

∴ \(V_0=\sqrt{\frac{G m}{R}}\), But \(g=\frac{G m}{R^2}, V_0=\sqrt{g R}\)

V₀ = \(\sqrt{gR}\) is called orbital velocity. Its value Is 7.92 km/sec.

Question 5. What Is escape velocity? Obtain an expression for It
Answer:

Escape Speed (V₁)_min: ft is defined as the minimum velocity required by a body to overcome the gravitational field of the earth is known as escape velocity.

For a body of mass ‘m’ gravitational potential energy on the surface of the earth

PE = \(-\frac{G \cdot m \cdot M_E}{R_E}\)

For a body to escape from the gravitational field of the earth its kinetic energy must be equal to or more than gravitational potential energy.

KE = \(\frac{1}{2} \mathrm{mv}^2\)

For minimum velocity \(\mathrm{KE}=\mathrm{PE}\)

∴ \(\frac{1}{2} m v^2=-\frac{G m \cdot M_E}{R_E} \cdot O R \quad v=\sqrt{\frac{2 G M_E}{R_E}}\)

∴ The minimum initial speed on the surface of the earth \(\left[v_i\right]_{\min }=\sqrt{\frac{2 G M_E}{R_E}}\) is called escape speed. (or) \(V=\sqrt{2 \mathrm{gR}_{\mathrm{E}}}\)

∴ \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\)

Question 6. What is a geostationary satellite? State its uses.
Answer:

A geostationary satellite will always appear to be stationary relative to Earth.

The time period of a geostationary satellite is equal to time period of rotation of the earth.

∴ Time period of geostationary orbit t = 24 hours. Satellites in geostationary orbit will revolve around the earth in the west to the east direction in an equatorial plane.

Uses Of Geostationary Satellites:

1. For the study of the upper layers of the atmosphere.
2. For forecasting the changes in atmosphere and weather.
3. To find the size and shape of the earth.
4. For investigating minerals and ores present in the earth’s crust.
5. For transmission of TV signals.
6. For the study of the transmission of radio waves.
7. For space research.

Question 7. If two places are at the same height from the mean sea level; One is a mountain and the other is in the air. At which place will ‘g’ be greater? State the reason for your answer.
Answer:

The ’g’ value on the mountain is greater than the ‘g’ value in the air even though both are at the same height.

For a point on a mountain, while deciding the ‘g’ value, the mass of the mountain is also considered which leads to change in the ‘g’ value depending on local conditions such as the concentration of huge mass at a particular place.

Whereas for a point in the air, no such effect is considered. Hence ‘g’ on the top of the mountain is more.

Question 8. The weight of an object is more at the poles than at the equator. At which of these can we get more sugar for the same weight? State the reason for your answer.
Answer:

If we are using common balance to measure sugar we will get some quantity of sugar both at the equator and at the poles.

Whereas if we are using spring balance to weigh sugar then the weight of sugar at poles is more. So we will get less quantity.

The weight of sugar at the equator is less. So we will get more quantity of sugar at the equator.

Question 9. If a nut becomes loose and gets detached from a satellite revolving around the Earth, will it fall down to Earth or will it revolve around Earth? Give reasons for your answer.
Answer:

If a nut is detached from a satellite revolving in the orbit then its velocity is equal to orbital velocity. So it continues to revolve in the same orbit. It does not fall to earth.

Question 10. An object projected with n velocity greater than or equal to 11.2 km s^-1 will not return to Earth. Explain the reason.
Answer:

The escape velocity of the earth Is 11.2 km/sec. If any body acquires a velocity of 11.2 km/ sec. or more its kinetic energy is more than gravitational potential energy.

So earth is not able to stop the motion of that body. So anybody with a velocity of 11.2 km/s or more will escape from the gravitational field of Earth and never come back to Earth.

Gravitation Solutions KSEEB Class 11 Physics Long Answer Questions

Question 1. Define gravitational potential energy and derive an expression for it associated with two particles of masses m₁ and m₂.
Answer:

The gravitational potential energy of a body at a point in the gravitational field of another. It is defined as the amount of work done in bringing the given body from infinity to that point in the field is called Gravitational potential energy.

Expression For Gravitational Potential Energy: Consider two particles of masses m₁ and m₂ placed at the points, ‘O’ and p respectively. Let the distance between the two particles be V i.e., OP = r.

Let us calculate the gravitational potential energy of the particle of mass m₂ placed at point p in the gravitational field of m₁ Join OP and extend it in the forward direction.

Consider two points A and B on this line such that OA = x and AB = dx.

The gravitational force of attraction on the particle at A is, \(\mathrm{F}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{x}^2}\)

A small amount of work done In bringing the particle without acceleration through a very small distance AB is, dW = F dx

= \(\mathrm{F}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{x}^2}\) dx

Total work done In bringing the particle from infinity to point P is,

N = \(\int_{\infty}^G \frac{G m_1 m_2}{x^2} d x=G m_1 m_2 \int_{\infty}^r x^{-2} d x\)

= \(-G m_1 m_2\left[\frac{1}{x}\right]_{\infty}^r=-G m_1 m_2\left[\frac{1}{r}-\frac{1}{\infty}\right]\)

= \(\frac{-G m_1 m_2}{r}\)

Since this work done is stored in the particle as its gravitational potential energy (U).

Therefore, the gravitational potential energy of the particle of mass m₂ placed at point ‘p’, in the gravitational field of a particle of mass m at distance r is \(\mathrm{U}=\frac{-G \mathrm{~m}_1 \mathrm{~m}_2}{} \text {. }\)

Here, a negative sign shows that the potential energy is due to an attractive gravitational force between two particles.

Question 2. Derive an expression for the variation of acceleration due to gravity

1. Above and
2. Below the surface of the Earth.

Answer:

Variation Of Acceleration Due To Gravity Above The Surface Of Earth: We know ‘g’ on the planet, g = \(\frac{G M}{R^2}\). But on earth ‘g’ value changes with height above the ground ‘h’

Variation Of ’g’ With Altitude: For a point ‘h’ above the earth, the total mass of the earth seems to be concentrated at the centre of the earth. Now the distance from the centre of the earth is (R + h).

Acceleration due to gravity at ‘h’ = g(h) = \(\frac{G M_E}{\left(R_E+h\right)^2}\)

For small values of ‘h’ l.e., h << R then \(g(h)=g\left(1-\frac{2 h}{R_E}\right)\)

Proof: \(g(h)=\frac{G M}{(R+h)^2}=\frac{G M}{R^2\left(1+\frac{h}{R}\right)^2}\) or

⇒ \(\mathrm{g}(\mathrm{h})=\frac{\mathrm{GM}}{\mathrm{R}^2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{-2}\)

(or) \(\mathrm{g}(\mathrm{h})=\mathrm{g}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{-2}\)

on Binomial expansion \(g(h)=g\left(1-\frac{2 h}{R}+\frac{4 h^2}{R^2} \ldots . . \text { etc. }\right)\).

By neglecting higher order terms, g(h) = \(g\left(1-\frac{2 h}{R}\right)\)

Variation Of Acceleration Due To Gravity Below The Surface Of Earth: At a depth d inside the groundmass of the earth up to point d from the centre will exhibit the force of attraction on the body. The remaining mass does not exhibit any influence.

Mass of spherical body M ∝ R³

∴ \(M_{/} / M_E=\left(R_E-d\right)^3 / R_E^3\) where \(M_s\) is mass of earth ‘s’ shell up to a depth ‘d’ from the centre. Gravitational force at depth ‘d’ is

F(d) = \(G M_s m /\left(R_E-d\right)^2=G M_E m\left(R_E-d\right) / R_E^2\)

∴ g(d) = \(\frac{F(d)}{m}=\frac{G M_E}{R_E^3}\left(R_E-d\right)=g\left(R_E-d\right)\)

= \(g\left(1-\frac{d}{R_E}\right)\)

So ‘g’ value decreases with depth below the ground.

Question 3. State Newton’s Universal Law of Gravitation. Explain how the value of the Gravitational constant (G) can be determined by the Cavendish method.
Answer:

Newton’s Law Of Gravitation: Everybody in the universe attracts every other body with a force which is directly proportional to the product of their maw”, and Inversely proportional to the square of the distance between them.

∴ \(F \propto m_1 m_2 \text { and } F \propto \frac{1}{d^2}\) or \(F \propto \frac{m_1 m_2}{d^2}\)

or \(F=\frac{G m_1 m_2}{d^2}\)

This is always a force of attraction and acts along the line joining the two bodies.

Cavendish Experiment To Find Gravitational Constant ‘G’: The Cavendish experiment consists of a long metallic rod AB to which two small lead spheres of mass ’m’ are attached.

This rod is suspended from a rigid support with the help of a thin wire. Two heavy spheres of mass M are brought near to these small spheres in opposite directions. Then gravitational force will act between the spheres.

Force between the spheres, \(\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{d}^2}\)

Two equal and opposite forces acting at the two ends of the rod AB will develop a force couple and the rod will rotate through an angle ‘θ’.

∴ Torque on the rod \(\mathrm{F} \times \mathrm{L}=\frac{\mathrm{GM} \cdot \mathrm{m}}{\mathrm{d}^2} \mathrm{~L}\)….(1)

Restoring force couple = τθ…(2)

Where τ = Restoring couple per unit twist

∴ \(\tau \theta=\frac{\mathrm{GMm}}{\mathrm{d}^2} \mathrm{~L}\)

or \(\mathrm{G}=\frac{\tau \cdot \theta \mathrm{d}^2}{\mathrm{MmL}}\)

By measuring we can calculate the “G” value when other parameters are shown.

The particle value of G is 6.67 x 10^-11 Nm²/kg²

KSEEB Class 11 Physics Chapter 8 Gravitation Problems

(Gravitational Constant ‘G’ = 6.67 x 10^-11 Nm²kg^-2, Radius of earth ‘R’ = 6400 km; Mass of earth M_E= x 10²⁴ kg)

Question 1. Two spherical halls each of mass 1 kg are placed 1 cm apart, a kind the gravitational force of attraction between them.
Solution:

Mass of each ball, m = 1 kg;

Separation, r = 1 cm = 10^-2 m

The gravitational force of attraction,

F = \(\frac{\mathrm{Gmm}}{\mathrm{r}^2}=\frac{6.67 \times 10^{-11} \times 1 \times 1}{\left(10^{-2}\right)^2}\)

= \(6.67 \times 10^{-11} \times 10^4=6.67 \times 10^{-7} \mathrm{~N}\)

Question 2. The mass of a ball is four times the mass of another ball. When these balls are separated by a distance of 10 cm, the gravitational force between them is 6.67 x 10^-7 N. Find the masses of the two balls.
Solution:

Mass of 1st ball = m;

Mass of 2nd ball = 4m.

Separation, r = 10 cm = 0.1 m;

Mass of the 1st ball = m =?

Gravitational force, F = 6.67 x 10^-7 N

Gravitational force, \(F=\frac{\mathrm{G} \cdot \mathrm{m} \cdot 4 \mathrm{~m}}{\mathrm{r}^2}\)

6.67 \(\times 10^{-7}=\frac{6.67 \times 10^{-11} 4 \mathrm{~m}^2}{0.1 \times 0.1}\)

∴ \(10^{-7}=10^{-9} .4 \mathrm{~m}^2\)

∴ \(\mathrm{m}^2=\frac{10^{-7}}{4 \times 10^{-9}}=\frac{100}{4}=25 \Rightarrow \mathrm{m}=5\)

∴ Mass of the balls is \(5 \mathrm{~kg}, 20 \mathrm{~kg}\).

Question 3. Three spherical balls of masses 1 kg, 2 kg and 3 kg are placed at the corners of an equilateral triangle of side 1 m. Find the magnitude of the gravitational force exerted by the 2 kg and 3 kg masses on the 1 kg mass.
Solution:

Side of an equilateral triangle, a = 1 m.

Masses at corners = 1 kg, 2 kg, 3 kg.

Force between 1 kg, 2g = \(F_1=G \frac{2 \times 1}{1^2}=2 G\).

Force between 1 kg, 3kg = \(F_2= G\frac{3 \times 1}{1^2}=3 \mathrm{G}\)

Now F₁ and F₂ act with an angle of 60″

∴ Resultant force \(F_K=\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos \theta}\)

= \(\sqrt{4 G^2+9 G^2+2 \times 2 \times 3 G^2 \times \frac{1}{2}}\)

= \(G \sqrt{4+9+6}=\sqrt{19} G\)

Question 4. At a certain height above the earth’s surface, the acceleration due to gravity is 4% of its value at the surface of the earth. Determine the height.
Solution:

Acceleration due to gravity at a height, h = 4% of g.

Radius of earth, R = 6400 K.M. = 6.4 x 10⁶ m

∴ \(g_h=\frac{4}{100} g . \quad B u t g_h=\frac{g \cdot R^2}{\left(1+\frac{h}{R}\right)^2}\)

∴ \(\frac{4}{100} g=\frac{g}{\left(1+h / R^2\right)} \Rightarrow\left(1+\frac{h}{R}\right)^2=\frac{100}{4}\)

Take roots on both sides then, \(+\frac{h}{R}=\frac{10}{2}\)

= \(5 \Rightarrow 1+\frac{\dot{h}}{R}=5 \Rightarrow \frac{h}{R}=5 .-1=4\)

∴ h = \(4 R=6400 \times 4=25,600 \mathrm{k} \cdot \mathrm{m}\)

Question 5. A satellite orbits the earth at a height of 1000 km. Find its orbital speed.
Solution:

Radius of earth, R = 6,400 km = 6.4 x 10⁶ m;

Mass of earth, M = 6 x 10²⁴

Height of satellite, h = 1000 km; G = 6.67 x 10¹¹ N-m²/kg²

Orbital velocity \(V_0=\sqrt{\frac{G M}{R+h}}\)

∴ \(\mathrm{V}_{\mathrm{o}}=\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6400+1000}}\)

= \(\sqrt{\frac{6.67 \times 6 \times 10^{13}}{7.4 \times 10^6}}=\sqrt{\frac{40.02 \times 10^7}{7.4}}\)

= 7354 m

Chapter 8 Gravitation Questions And Answers KSEEB Physics

Question 6. A satellite orbits the Earth at a height equal to the radius of Earth. Find it’s

1. Orbital speed and
2. Period of revolution.

Solution:

Radius of earth, R = 6400 k.m.;

height above the earth, h = R.

Mass of earth, M = 6 x 10²⁴;

G = 6.67 x 10^-11 N – m²/kg²

1. Orbital velocity, \(\mathrm{V}_0=\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}\)

∴ \(\mathrm{V}_0=\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{2 \times 6.4 \times 10^6}}\)

= \(\sqrt{\frac{40.02 \times 10^{13}}{12.8 \times 10^6}}=\sqrt{\frac{40.02 \times 10^7}{12.8}}\)

= \(5592 \mathrm{~m} / \mathrm{s}=5.592 \mathrm{~km} / \mathrm{sec}\)

2. Time period, \(\mathrm{T}=\frac{2 \pi(2 \mathrm{R})}{\mathrm{V}}\)

= \(\frac{2 \times 3.142 \times 6.4 \times 10^6 \times 2}{5592}=14,380 \mathrm{sec}\)

= \(3.994 \mathrm{~h} \approx 4 \text { hours. }\)

Question 7. The gravitational force of attraction between two objects decreases by 36% when the distance between them is increased by 4 m. Find the original distance between them.
Solution:

Let force between the objects = F;

Distance between them = r.

For Case II distance, r₁ = (r + 4);

The new force, F₁ = 36% less than F

∴ \(F_1=F\left(1-\frac{36}{100}\right)=\frac{64}{100} F\)

⇒ \(\frac{\text { G.m.m. }}{(r+4)^2}=\frac{64}{100} \frac{\text { G.m.m }}{r^2}\)

⇒ \(100 r^2=64(r+4)^2\) Take square roots on both sides

⇒ \(10 \mathrm{r}=8(\mathrm{r}+4) \Rightarrow 10 \mathrm{r}=8 \mathrm{r}+32\)

∴ (10-8)r = 32

∴ \(\mathrm{r}=16 \dot{\mathrm{m}} \text {. }\)

Question 8. Four identical masses m are kept at the corners of a square of side a. Find the gravitational force exerted on one of the masses by the other masses.
Solution:

Given all masses are equal m₁ = m₂ = m₃ = m₄

Force between \(m_1, m_4=F_1=\frac{G \cdot m^2}{a^2}\)….(1)

Force between \(m_4, m_3=F_2=\frac{G \cdot m^2}{a^2}\)…….(2)

Forces F₁ and F₂ act perpendicularly.

Their magnitudes are equal.

∴ \(F_R=\sqrt{2 F}=\sqrt{2} \cdot \frac{\mathrm{Gm}^2}{\mathrm{a}^2}\)…..(3) (From Parallelogram Law)

Force between \(m_4, m_2=\frac{G m^2}{(\sqrt{2 a})^2}=\frac{G m^2}{2 a^2}\)

(say \(\mathrm{F}_3\)) ….. (4)

Now forces \(F_R\) and \(F_3\) are like parallel. So resultant is the sum of these forces.

Total force at \(\mathrm{m}_4\) due to other masses

= \(\sqrt{2} \frac{G m^2}{a^2}+\frac{G m^2}{2 a^2}=\frac{G m^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)\)

Question 9. Two spherical balls of 1 kg and 4 kg are separated by a distance of 12 cm. Find the distance of a point from the 1 kg mass at which the gravitational force on any mass becomes zero.
Solution:

Mass, m₁ = 1 kg ;

Mass, m₂ = 4 kg ;

Separation, d = 12 cm

Mass of 3rd body m₃ =?

For m₃ not to experience any force the condition is

Force between m₁, m₃ = Force between

∴ \(\frac{G_1 \times m_3}{x^2}=\frac{G \times 4 m_3}{(d-x)^2} \Rightarrow(d-x)^2=4 x^2\)

Take square roots on both sides,

d-x = \(2 x \Rightarrow d=3 x \text { or } x=\frac{12}{3}=4 \mathrm{~cm}\)

∴ Distance from 1 kg mass = 4 cm

Question 10. Three uniform spheres each of mass m and radius R are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.
Solution:

Mass m and radius R are the same for all spheres.

Force between 1, 3 spheres = \(F_1=\frac{G \cdot m^2}{(2 R)^2}\)

Force between 1, 2 spheres = \(F_2=\frac{G \cdot m^2}{(2 R)^2}\)

Now F₁ and F₂ will act with an angle θ = 60° between them so from Parallelogram law

Resultant force F = \(\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos \theta}\)

= \(\sqrt{F^2+F^2+2 F^2 \frac{1}{2}}=\sqrt{3} F\)

∴ Force on 1st sphere due to other two spheres \(=\frac{\sqrt{3} \cdot \mathrm{Gm}^2}{4 \mathrm{R}^2}\)

Force on 1st sphere due to other two spheres = \(\frac{\sqrt{3} \cdot \mathrm{Gm}^2}{4 \mathrm{R}^2}\)

Question 11. Two satellites are revolving around the earth at different heights. The ratio of their orbital speeds is 2:1. If one of them is at a height of 100 km what is the height of the other satellite?
Solution:

Mass of the earth, m = 6 x 10²⁰ kg

G = 6.67x 10^-11N-m²/kg²

The ratio of orbital velocities V₀₁: V₀₂ = 2:1;

Height of one satellite, h = 100 k.m

But \(\mathrm{V}_0=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}\)

∴ \(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}_1}}=\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}_2}}\)

By squaring on both sides \(\frac{\mathrm{Gm}}{\mathrm{R}+\mathrm{h}_1}=\frac{1}{4} \frac{\mathrm{Gm}}{\mathrm{R}+\mathrm{h}_2}\)

⇒ \(4\left(R+h_2\right)=R+h_1\)

⇒ \(4 R+4 h_2=R+h_1 \Rightarrow h_1=3 R+4 h_2\)

Put \(\mathrm{h}_2=100 \mathrm{~km}\)

∴ \(\mathrm{h}_1=3 \times 6400+400=19600 \mathrm{~km} \text {. }\)

Question 12. A satellite revolves around in a circular orbit with a speed of 8 km/s^-1 at a height where the value of acceleration due to gravity is 8 m/s^-2. How high is the satellite from the Earth’s surface? (Radius of the planet 6000 km.).
Solution:

Orbital velocity of satellite, V₀ =8km/s. = 8 x 10³ m/s.

Acceleration due to gravity in the orbit = g = 8 m/s²

Orbital Velocity \(V=\sqrt{g R}\)

Where R is the radius of the orbit and g is the acceleration due to gravity In the orbit.

∴ \(R=V^2 / g=\frac{8 \times 8 \times 10^6}{8}=8 \times 10^6 \mathrm{~m}\)

= \(8000 \mathrm{~km}\)

Height of satellite = 8000 – radius of earth; Radius of earth = 6000 km.

∴ Height above earth = 8000 – 6000 = 2000 km.

Question 13.

1. Calculate the escape velocity of a body from the Earth’s surface,
2. If the Earth were made of wood, its mass would be 10% of its current mass. What would be the escape velocity, if the Earth were made of wood?

Solution:

The radius of the earth, R = 6400 km = 6.4 x 10⁶ m.

Mass of earth, M = 6 x 1024 kg; g = 9.8 ms^-2.

1. Escape velocity, \(\mathrm{V}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\)

∴ \(\mathrm{V}_{\mathrm{e}}=\sqrt{2 \times 9.8 \times 6.4 \times 10^6}=11.2 \mathrm{~km} / \mathrm{s}\)

2. If the earth is made of wood mass, \(M_1=10 \% \text { of } M=6 \times 10^{23}\)

Escape velocity, \(V_e=\sqrt{\frac{2 G m}{R}}\)

= \(\sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6 \times 10^{23}}{6.4 \times 10^6}}\)

= \(\sqrt{\frac{2 \times 40.02 \times 10^{12}}{6.4 \times 10^6}}\)

∴ \(\mathrm{V}_{\mathrm{e}} =\sqrt{12.51 \times 10^6}=3.537 \mathrm{~km} / \mathrm{s}\)

Question 14. A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant

1. Linear speed
2. Angular speed
3. Angular momentum
4. Kinetic energy
5. Potential energy
6. Total energy throughout its orbit?

Neglect any mass loss of the comet when it comes very close to the Sun.

Solution:

A comet while going on elliptical orbit around the Sun has constant angular momentum and total energy at all locations but other quantities vary with locations.

Question 15. A Saturn year is 29.5 times the Earth year. How far is Saturn from the sun if the Earth is 1.5 x 10⁸ km away from the sun?
Solution:

Here, T_s = 29.5 T_e; R_e = 1.5 x 10⁸ km; R_s = ?

Using the relation, \(\frac{T_{\mathrm{s}}^2}{\mathrm{R}_{\mathrm{s}}^3}=\frac{\mathrm{T}_{\mathrm{e}}^2}{\mathrm{R}_{\mathrm{e}}^3}\)

or \(\mathrm{R}_{\mathrm{s}}=\mathrm{R}_{\mathrm{e}}\left(\frac{\mathrm{T}_{\mathrm{s}}}{\mathrm{T}_{\mathrm{e}}}\right)^{2 / 3}\) = \(1.5 \times 10^8\left(\frac{29.5 \mathrm{~T}_{\mathrm{e}}}{\mathrm{T}_{\mathrm{e}}}\right)^{2 / 3}\)

= \(1.43 \times 10^9 \mathrm{~km}\)

KSEEB Physics Class 11 Chapter 8 Gravitation Notes

Question 16. A body weighs 63 N on the surface of Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Solution:

Weight of body = mg = 63 N

At height h, the value of ‘g’ is given by, \(g^{\prime}=\frac{g R^2}{(R+h)^2}=\frac{g R^2}{(R+R / 2)^2}=\frac{4}{9} g\)

The gravitational force on the body at a height of h is

⇒ \(\mathrm{F}=\mathrm{mg}^{\prime}=\mathrm{m} \times \frac{4}{9} \mathrm{~g}=\frac{4}{9} \mathrm{mg}=\frac{4}{9} \times 63=28 \mathrm{~N}\)

Question 17. Assuming the earth to be a sphere of uniform mass density, how much would the body weigh halfway down to the centre of the earth if it weighed 250 N on the surface?
Solution:

Weight of body at a depth, d = mg’

= \(m \times g\left(1-\frac{d}{R}\right)=250\left(1-\frac{\frac{R}{2}}{R}\right)=125 \mathrm{~N}\)

KSEEB Solutions For Class 11 Physics Chapter 14 Oscillations Important Points

Periodic Motion: A motion that replies itself at regular intervals of time is called periodic motion.

Oscillations Or Vibrations: When a small displacement is given to a body at rest position (i.e. its mean position) then a force will come into play and try to bring back the body to its rest position or equilibrium position by executing and ho motion about a mean position. These are called vibrations or oscillations.

Time period (T): In periodic motion, the smallest time interval after which the motion is repeated is called its time period.

Displacement: For a body in periodic motion the displacement changes with time, so displacement is given by x(t) (or) f(t) = A cos ωt

Frequency: The reciprocal of time period (T) gives the number of repetitions that occur for unit time. It is called frequency υ).

υ = \(\frac{1}{T}\) = 1/Time period

Read and Learn More KSEEB Class 11 Physics Solutions

Fourier Theory: According to Fourier “any periodic function can be expressed as a superposition of sine and cosine functions of different time periods with suitable coefficients.”

Simple Harmonic Motion (Explanation Of Equation): In simple harmonic motion displacement is a sinusoidal function of time. i.e. x(t) = Acos(ωt±ø)

The particle will oscillate back and forth about the origin on the X-axis within the limits +A and -A where A is amplitude, ω and ø are constants.

1. Amplitude A of simple harmonic motion (S.H.M.) is the magnitude of the maximum displacement of the particle. Displacement of particles varies from +A to -A.
2. Phase Of Motion (Or) Argument: For a body in S.H.M the position of a particle or state of motion of a particle at any time V is determined by the argument (wt + ø). The term (wt + ø) is called argument or phase of motion.
3. Phase Constant (Or) Phase Angle: The value of the phase of motion at time t=0 is called “phase constant ø” or “phase angle”.
4. Simple harmonic motion is represented with a cosine function. It has a periodicity of 2π. The function repeats after a time period T.

Reference Circle: Let a particle ’p’ move uniformly on a circle. The projection of p on any diameter of the circle will execute S.H.M.

The particle ‘p’ is called a “reference particle” and the circle on which the particle moves is called a “reference circle”.

Velocity Of A Body On S.H.M.: For a body in uniform circular motion speed v = ωA. The direction of velocity \(\bar{V}\), at any time is along the tangent to the circle at that instantaneous point.

Mathematically velocity of the body \(\bar{V}\)(t) = — ωAsin(ωt + ø)

Or, \(v(t)=\frac{d}{d t} x(t)=\frac{d}{d t} A \cos (\omega t+\phi)\)

= \(-\omega \mathrm{A} \sin (\omega t+\phi)\)

KSEEB Class 11 Physics Solutions For Chapter 14 Oscillations

Acceleration (a): For a body In S.H.M the Instantaneous acceleration of the particle is a(t) = -ω² A cos(wt + ø) = -ω²x(t)

or \(a(t)=\frac{d}{d t} v(t)=\frac{d}{d t}[-A \omega \sin (\omega t+\phi)]\)

= \(-A \omega^2 \cos (\omega t+\phi)\)

a(t) implies the acceleration of the body is a function of time.

Maximum acceleration of the body a_max = -ωA.

The -ve sign indicates that acceleration and displacement are in opposite directions.

Force On A Body In S.H.M: Acceleration a(t) = – ²x(t)

Force F = m a

∴ Force on a body in S.H.M

F(t) = ma = m(-ω²xt) = – K x(t)

Where K = mω² and m = mass of the body

Energy Of A Body In S.H.M: For a body in S.H.M both kinetic energy and potential energy will change with time. These values will vary between zero and their maximum value.

Kinetic energy K  = \(\frac{1}{2} \mathrm{mv}^2\)

= \(\frac{1}{2} \mathrm{m \omega}^2 \mathrm{~A}^2 \sin ^2(\omega t+\phi)\)

= \(\frac{1}{2} \mathrm{kA}^2 \sin ^2(\omega t+\phi)\)

Potential energy U = \(\frac{1}{2} \mathrm{kx}^2\)

= \(\frac{1}{2} \mathrm{kA}^2 \cos ^2(\omega t+\phi)\)

Potential energy \(\mathrm{U}=\frac{1}{2} \mathrm{kx}^2\)

Springs:

1. In the case of springs for smaller displacements when compared with the length of the spring Hooke’s law will hold good.
2. The small oscillations of a block of mass ‘m’ connected to a spring can be taken as simple harmonic.
3. In the case of springs the restoring force acting on the block of mass m is F(x) = -k(x)
4. Spring constant k is defined as the force required for unit elongation. Unit: newton/meter, K= force/displacement ⇒ K = \(\frac{-F}{x}\) -ve sign indicates that x force and displacement are opposite in direction. For a stiff spring, k is high. For a soft spring, k is less.
5. Angular frequency of a loaded spring \(\omega=\sqrt{\frac{K}{m}}\)

Time period of pendulum = \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)

Where m is the load attached and k is the constant of spring.

Simple Pendulum: Time period of oscillation \(\mathrm{T}=2 \pi \sqrt{l / g}\)

Damped Oscillations: In damped oscillations, the energy of the system is dissipated continuously.

When damping is very small the oscillations will remain approximately periodic.

Damped Oscillations Example: Oscillations of a simple pendulum.

Damping force depends on the nature of the medium.

Free Oscillations: When a system is displaced from its equilibrium position and allowed free to oscillate then oscillations made by the body are known as free oscillations.

The frequency of vibration is known as the natural frequency of that body.

Forced Or Driven Oscillations: If a body is made to oscillate under the influence of an external periodic force (say ω₀) then those oscillations are called forced oscillations.

Resonance: The phenomenon of an increase in the amplitude of the vibrating body when driving force frequency ‘ω₀‘ is equal to or very close to the natural frequency ‘ω’ of the oscillator is called resonance.

When the driving frequency is very close to the natural frequency of the vibrating body then also resonance will occur. Due to this reason damage to buildings is caused in earthquake-affected areas.

KSEEB Class 11 Physics Chapter 14 Oscillations Important Formulae

Displacement of a body in S.H.M Y = A sin (ωt ± ø)

The velocity of a body in S.H.M is

V = \(\frac{d y}{d t}=\frac{d}{d t}(A \sin \omega t)\)

V = \(A \omega \cos \omega t\) where \(Y=A \sin \omega t\)

or \(V=\omega \sqrt{A^2-Y^2}\)

Maximum velocity, \(\mathrm{V}_{\max }=\mathrm{A} \omega\)

Acceleration of a body in S.H.M a = – ω² A sin ωt or a = – ω²Y (where Y = A sin ωt)

(The -ve sign shows that acceleration and displacement are opposite to each other)

Maximum acceleration, a_max = ω²A

Angular Velocity ‘ω’ Of A Body In S.H.M: a ∝ -Y or a = -ω²Y (-ve sign for opposite direction only)

∴ \(\omega^2=\frac{a}{Y}\) or \(\omega=\sqrt{\frac{a}{Y}}\)

∴ Angular velocity of a body in S.H.M, \(\omega=\sqrt{\frac{\text { acceleration }}{\text { displacement }}}\)

Time Period Of A Body In S.H.M: The time taken for one complete oscillation is called the “time period”.

Time period, (T) = \(\frac{\text { Angular displacement for one rotation }}{\text { Angular velocity }}=\frac{2 \pi}{\omega}\)

∴ T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{Y}}{\mathrm{a}}}=2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}}\)

Frequency (ν) is the number of vibrations (or) rotations per second.

Frequency, v = \(\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{a}{Y}}\)

⇒ v = \(\frac{\omega}{2 \pi}\) or \(\omega=2 \pi v\)

Oscillations Solutions KSEEB Class 11 Physics

Springs:

1. In springs force constant is defined as the ratio of Force to displacement.

Spring constant, (K) = \(\frac{F}{Y}\)

2. Time period, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)

3. If the mass of the spring m₁ is also taken into account, then the time period, \(T_1=2 \pi \frac{\sqrt{\left(m+\frac{m_1}{3}\right)}}{K}\) (For real spring )

4. Frequency of vibration, n = \(\frac{1}{2 \pi} \sqrt{\frac{K}{m}}\)

For real spring, n = \(\frac{1}{2 \pi} \sqrt{\frac{K}{\left(m+\frac{m_1}{3}\right)}}\)

When a spring is divided into ‘n’ parts then its force constant (k) will increase.

New force constant k₁ = nk

where n = number of parts and k = original spring constant.

Simple Pendulum:

1. In a simple pendulum component of the weight of the bob useful for to and motion is F = mg sin θ
2. Time period, \(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \Rightarrow \mathrm{g}=4 \pi^2 \frac{l}{\mathrm{~T}^2}\)
3. When a simple pendulum is placed in a lift moving with some acceleration then its time period chants.
4. When lift moves up with an acceleration its time period decreases, \(T=2 \pi \sqrt{\frac{1}{g+a}}\)
5. When lift moves down with an acceleration ‘a’ its lime period increases, \(\mathrm{T}=2 \pi \sqrt{\frac{l}{g-\mathrm{a}}}\)
6. For a seconds pendulum time period, T = 2 sec
7. Length of seconds pendulum on earth = 100 cm = 1 m ( nearly )
8. In a simple pendulum L – T² graph is a straight line passing through the origin.

KSEEB Physics Class 11 Oscillations Very Short Answer Questions

Question 1. Give two examples of periodic motion which are not oscillatory.
Answer:

1. Motion of seconds hand of a watch.
2. Motion of fan blades which are rotating with constant angular velocity ‘w’.

For these two cases, they have constant centrifugal acceleration which does not change with rotation so it is not considered as S.H.M.

Question 2. The displacement in S.H.M. is given by y = a sin (20t + 4). What is the displacement when it is increased by 2π/ω?
Answer:

Displacement: Displacement remains constant; \(\frac{2 \pi}{\omega}\) = time period T.

After a time period of T, there is no change in the equation of S.H.M

i.e. Y = A sin (20t + 4) = Y  = A sin (20 t + 4 + T)

∴ There is no change in displacement.

Question 3. A girl is swimming seated in a swing. What is the effect on the frequency of oscillation if statement?
Answer:

The frequency of oscillation (n) will Increase because, in the standing position, the location of the center of mass of the girl shifts upwards. Due to this, the effective length of the swing decreases.

As increases \(\mathrm{n} \propto \frac{1}{\sqrt{1}}\), therefore, n increases.

Question 4. The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hollow sphere?
Answer:

When water begins to drain out of the sphere, the center of mass of the system will first move down and then will come up to the initial position. Due to this the equivalent length of the pendulum and hence time period first increases, reaches a maximum value, and then decreases till it becomes equal to its initial value.

Question 5. The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminum?
Answer:

The time period of a simple pendulum does not change if the wooden bob is replaced by an identical bob of aluminum because the time period of a simple pendulum is independent of the material of the bob.

Question 6. Will a pendulum clock gain or lose time when taken to the top of a mountain?
Answer:

At higher altitudes i.e., on mountains, the acceleration due to gravity is less as compared to the surface of the earth. Since the period is inversely proportional to the square root of the acceleration due to gravity, the time period increases. The pendulum clock loses time on the top of a mountain.

KSEEB Physics Class 11 Oscillations Notes

Question 7. What is the length of a simple pendulum which ticks seconds? (g = 9.8 ms^-2)
Answer:

In simple pendulum \(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \text { or } l=\frac{\mathrm{gt}^2}{4 \pi^2}\)

For seconds pendulum T = 2s

⇒ t² = 4

∴ l = \(\frac{9.8 \times 4}{4 \pi^2}=1 \mathrm{~m}\) (π² nearly 9.8)

Question 8. Wind happens to The time period of n simple pendulum If Its length is increased up to four times.
Answer:

In a simple pendulum Time period T ∝ √l

∴ \(\frac{T_1}{T}=\sqrt{\frac{l_1}{l}}\) given \(l_1=4 l\)

∴ \(T_1=T \sqrt{\frac{4 l}{l}}=2 \mathrm{~T}\)

From the above equation, the period is doubled.

Question 9. A pendulum clock gives the correct time at the equator. Will it gain or lose time if it is taken to the poles? If so, why?
Answer:

When a pendulum clock showing the correct time at the equator is taken to poles then it will gain time.

Acceleration due to gravity at the poles is high. Time period of pendulum \(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\)

When g increases T decreases. So several oscillations made in the given time increase hence clock gains time.

Question 10. What fraction of the total energy is K.E when the displacement is one-half of the amplitude of a particle executing S.H.M?
Answer:

Kinetic energy is equal to the fourth (i.e., \(\frac{3}{4}\)) of the total energy when the displacement is one-half of its amplitude.

Question 11. What happens to the energy of a simple harmonic oscillator if its amplitude is doubled?
Answer:

The energy of a simple harmonic oscillator,

E = \(\frac{1}{2} m \omega^2 A^2\)

E \(\propto A^2 \Rightarrow \frac{E_1}{E_2}=\frac{A_1^2}{A_2^2} ; E_2=\frac{A_2^2}{A_1^2} \times E_1\);

⇒ \(E_2=\frac{4 A^2}{A^2} \times E_1 \Rightarrow E_2=4 E_1\)

From the above equation energy increases by four lines.

Question 12. Can a simple pendulum be used In an artificial satellite? Give the reason.
Answer:

No, this Is because, Inside the satellite, there is no gravity, i.e., g = 0. As \(\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}\), where T = ∞ for g = 0. Thus, the simple pendulum will not oscillate.

KSEEB Class 11 Physics Chapter 14 Oscillations Short Answer Questions

Question 1. Define simple harmonic motion. Give two examples.
Answer:

Simple Harmonic Motion: A body is said to be in S.H.M. if its acceleration is directly proportional to its displacement, and acts opposite in direction towards a faced point.

Examples:

1. Projection of uniform circular motion on a diameter.
2. Oscillations of a simple pendulum with| small amplitude.
3. Oscillations of a loaded spring.
4. Vibrations of a liquid column in a U-tube.

Question 2. Present graphically the variations of displacement, velocity, and acceleration with time for a particle in S.H.M.
Answer:

The variations of displacement, velocity, and acceleration with time for a particle in S.H.M can be represented graphically as shown) in the figure.

From The Graph

1. All quantities vary sinusoidally with time.
2. Only their maxima differ and the different plots differ in phase.
3. Displacement x varies between – A to A; v(t) varies from -ωA to ωA and a(t) varies from – ω²A to ω²A.
4. With respect to the displacement plot, the velocity plot has a phase difference of \({\pi}{2}\), and the acceleration plot has a phase difference of π.

Question 3. What is phase? Discuss the phase relations between displacement, velocity, and acceleration in simple harmonic motion.
Answer:

Phase (θ): Phase is defined as its state or condition as regards its position and direction of motion at that instant.

In S.H.M phase angle, \(\theta=\omega t=2 \pi\left(\frac{t}{T}\right)\)

1. Phase Between Velocity And Displacement: In S.H.M, displacement,

y = \(A \sin (\omega t-\phi)\)

Velocity, \(V=A \omega \cos (\omega t-\phi)\)

So phase difference between displacement and velocity is 90°.

2. Phase Between Displacement And Acceleration:

In S.H.M, acceleration ‘a’ = – ω² y

or y = A sin ωt and a = – ω²A sin ωt

The -ve sign indicates that acceleration and displacement are opposite.

So phase difference between displacement and acceleration is 180°.

KSEEB Class 11 Physics Oscillations Key Concepts

Question 4. Obtain an equation for the frequency of oscillation of spring of force constant k to which a mass m is attached.
Answer:

Let a spring of negligible mass is suspended from a fixed point and mass m is attached as shown. It is pulled down by a small distance x and allowed free it will execute simple harmonic oscillations.

Displacement from mean position = x

The restoring forces developed are opposite to displacement and proportional to ‘x’.

∴ F ∝ -x or F = -kx where k is constant of spring, (-ve sign for opposite direction)

Acceleration \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{Kx}}{\mathrm{m}} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{m}}{\mathrm{K}}\)

Time period \(\mathrm{T}=2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}}=2 \pi \sqrt{\frac{\mathrm{x}}{\mathrm{a}}}\)

For spring \(\frac{\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{m}}{\mathrm{K}}\)

∴ \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)

Time period of loaded spring \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)

Question 5. Derive expressions for the kinetic energy and potential energy of a simple harmonic oscillator.
Answer:

Expression For K.E Of A Simple Harmonic Oscillator: The displacement of the body in S.H.M., X = A sin ωt

where A = amplitude, ωt = Angular displacement.

Velocity at any instant, v = \(\frac{d x}{d t}\) Aω cos ωt

∴ K.E = \(\frac{1}{2} m v^2=\frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t\)

∴ \(K \cdot E=\frac{1}{2} m A^2 \omega^2\left(1-\sin ^2 \omega t\right)\)

= \(\frac{1}{2} m A^2 \omega^2\left[1-\frac{x^2}{\mathrm{~A}^2}\right]\)

= \(\frac{1}{2} m \omega^2\left[\mathrm{~A}^2-\mathrm{x}^2\right]\)

∴ \(K. E=\frac{1}{2} m \omega^2\left[\mathrm{~A}^2-\mathrm{x}^2\right]\)

At mean position, velocity is maximum, and displacement x=0

∴ \(\mathrm{K}_{\mathrm{max}}=\frac{1}{2} \mathrm{~mA}^2 \omega^2\)

Expression For P.E Of A Simple Harmonic Oscillator: Let a body of mass ‘m’ be in S.H.M with an amplitude A.

Let O be the mean position.

The equation of a body in S.H.M is given by, x = A sin ωt

For a body in S.H.M acceleration, a = – ω²Y

Force, F = ma = – mω²x

∴ Restoring force, F = mω² x

Potential Energy Of The Body At Any Point Say ‘x’: Let the body be displaced through a small distance dx

Work done, dW = F • dx = P.E.

This work is done.

∴ P.E = mω²x • dx(where x is its displacement)

Total work done, \(\mathrm{W}=\int \mathrm{dW}=\int_0^{\mathrm{x}} \mathrm{m} \omega^2 \mathrm{x} \cdot \mathrm{dx}\)

⇒ Work done, \(\mathrm{W}=\frac{\mathrm{m} \omega^2 \mathrm{x}^2}{2}\)

This work is stored as potential energy.

∴ P.E at any point = \(\frac{1}{2} m \omega^2 x^2\)

Oscillations Chapter 14 Questions And Answers KSEEB Physics

Question 6. How does the energy of a simple pendulum vary as it moves from one extreme position to the other during its oscillations?
Answer:

The total energy of a simple pendulum is, \(\mathrm{E}=\frac{1}{2} \mathrm{~mA}^2 \frac{(\mathrm{g})}{l}\) (or) \(\mathrm{E}=\frac{1}{2} \frac{\mathrm{mg}}{l} \mathrm{~A}^2\)

The above equation shows that the total energy of a simple pendulum remains constant irrespective of the position at any time during the oscillation i.e., the law of conservation of energy is valid in the case of a simple pendulum.

At the extreme positions, P and Q the energy is completely in the form of potential energy, and at the mean position O it is totally converted as kinetic energy.

At any other point, the sum of the potential and kinetic energies is equal to the maximum kinetic energy at the mean position or maximum potential energy at the extreme position.

As the bob of the pendulum moves from P to O, the potential energy decreases but appears in the same magnitude as kinetic energy.

Similarly, as the bob of the pendulum moves from O to P or Q, the kinetic energy decreases to the extent it is converted into potential energy, as shown in the figure.

Question 7. Derive the expressions for displacement, velocity, and acceleration of a particle executes S.H.M.
Answer:

Displacement of a body in S.H.M.

X = A cos (ωt + ø).

Displacement (x): At t = 0 displacement x = A i.e., at the extreme position when ωt + ø = 90° displacement x = 0 at a mean position at any point x = A cos (ωt + ø).

Velocity (V): Velocity of a body in S.H.M. is V = \(\frac{d x}{d t}=\frac{d}{d t}\{A \omega \cos (\omega t+\phi)\}\)

V = \(-A \omega \sin (\omega t+\phi)\) or

∴ V = \(-\omega \sqrt{A^2-x^2}\)

When (ωt + ø) = 0 then velocity v = 0. For points where (ωt + ø) = 90°

Velocity V = – Aω i.e., velocity is maximum

Acceleration (a): Acceleration of a body in S.H.M. is a = \(\frac{dv}{dx}\)

= \(\frac{d}{d t}(-A \omega \sin (\omega t+\phi)\)

= \(-A \omega^2 \cos (\omega t+\phi)=-\omega^2 x\)

∴ \(a_{\max }=-\omega^2 A\)

Chapter 14 Oscillations Long Questions And answers KSEEB Physics

Question 1. Define simple harmonic motion. Show that the motion of projection of a particle performing uniform circular motion, on any diameter is simple harmonic.
Answer:

Simple Harmonic Motion: A body is said to be in S.H.M. if its acceleration is directly proportional to its displacement, and acts opposite in the direction towards a fixed point.

Relation between uniform circular motion and S.H.M.: Let a particle ‘P’ rotate in a circular path of radius ‘r’ with a uniform angular velocity ‘ω’. After time’ it goes to a new position ‘P’.

Draw normals from ‘P’ onto the X-axis and onto the Y-axis. Let ON and OM be the projections on the X and Y axes respectively.

As the particle Is In motion it will subtend an angle θ = ωt at the centre.

From triangle OPN

ON = OP cos θ

But OP = r and θ = ωt

∴ Displacement of particle P on X – axis at any time t is X = r cos ωt…..(1)

From triangle OPM

OM = Y = OP sin θ

But OP = r and θ = ωt

∴ Displacement of particle P on Y- axis is Y = r sin ωt……(2)

As the particle rotates in a circular path the foot of the perpendiculars OM and ON will oscillate within the limits X to X¹ and Y to Y¹.

At any point, the displacement of particle P is given by OP² = OM² + ON²

Since OM = X = r cos ωt and ON = Y = r sin ωt.

So a uniform circular motion can be treated as a combination of two mutually perpendicular simple harmonic motions.

Question 2. Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is a second pendulum?
Answer:

Simple Pendulum: A massive metallic bob is suspended from a rigid support with the help of an inextensible thread. This arrangement is known as a simple pendulum.

So the length of a simple pendulum is T. Let the pendulum be pulled to a side by a small angle ‘θ’ and released it oscillate about the mean position.

Let the bob be at one extreme position B. The weight (W = mg) of the body acts vertically downwards.

By resolving the weight into two perpendicular components:

One component mg sin θ is responsible for the to and fro motion of the pendulum.

Other components mg cos θ will balance the tension in the string.

Force useful for motion F = mg sin θ = ma (From Newton’s 2nd Law)

From the above equations a = g sin θ

When is small \(\sin \theta \approx \theta=\frac{\text { arc }}{\text { radius }}=\frac{x}{l}\)

∴ a = \(g \frac{x}{l} \ldots \ldots \ldots \ldots . .(1) \Rightarrow a \propto x\)

Since acceleration is proportional to displacement and acceleration is always directed towards a fixed point the motion of the simple pendulum is “simple harmonic”. Period of the simple pendulum:

Time period of pendulum = \(2 \pi \sqrt{\frac{Y}{a}}\)

= \(2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}}\)

From equation (1) \(\mathrm{a}=\frac{\mathrm{g}}{l} \mathrm{x}\)

or \(\frac{\mathrm{x}}{\mathrm{a}}=\frac{\text { displacement }(\mathrm{y})}{\text { acceleration }(\mathrm{a})}=\frac{l}{\mathrm{~g}}\)

∴ Time period of simple pendulum, \(\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}\)

Seconds Pendulum: A pendulum whose time period is 2 seconds is called “seconds pendulum.”

KSEEB Class 11 Physics Chapter 14 Oscillations

Question 3. Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path.
Answer:

Expression for K.E of a simple harmonic oscillator: The displacement of the body in S.H.M, X = A sin ωt

where A = amplitude and ωt = Angular displacement.

Velocity at any instant, v = \(\frac{dx}{dt}\) = Aω cos ωt

So its K.E = \(\frac{1}{2} m v^2=\frac{1}{2} m A^2 \omega^2 \cos ^2 \omega \mathrm{t}\)

But \(\cos ^2 \omega t=\left(1-\sin ^2 \omega t\right)\)

∴ \(\mathrm{K} . E=\frac{1}{2} \mathrm{~mA}^2 \omega^2\left(1-\sin ^2 \omega \mathrm{t}\right)\)

= \(\frac{1}{2} m A^2 \omega^2\left[1-\frac{\mathrm{x}^2}{\mathrm{~A}^2}\right]\)

= \(\frac{1}{2} m \omega^2\left[A^2-x^2\right]\)

∴ \(\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^2\left[\mathrm{~A}^2-\mathrm{x}^2\right]\)

At mean position, velocity is maximum, and displacement x=0

∴ \(\mathrm{K} \cdot \mathrm{E}_{\max }=\frac{1}{2} \mathrm{~mA}^2 \omega^2\)

Expression for P.E Of A Simple Harmonic Oscillator: Let a body of mass ‘m’ be in S.H.M with an amplitude A.

Let O be the mean position.

The equation of a body in S.H.M is given by, x = A sin ωt

For a body in S.H.M acceleration, a = – ω²Y

Force, F = ma = – mω²x

∴ Restoring force, F = mω²x

Potential Energy Of The Body At Any Point Say ‘x’: Let the body be displaced through a small distance dx

⇒ Work done, dW = F • dx

This work done = P.E. in the body

∴ P.E = mω²x • dx(where x is its displacement)

Total work done, \(\mathrm{W}=\int \mathrm{dW}=\int_0^{\mathrm{x}} \mathrm{m} \omega^2 \mathrm{x} \cdot \mathrm{dx}\)

work done, \(\mathrm{W}=\frac{\mathrm{m} \omega^2 \mathrm{x}^2}{2}\).

This work is stored as potential energy.

∴ P.E at any point \(=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2\)

This work is stored as potential energy.

P.E at any point = \(\frac{1}{2} m \omega^2 x^2\)

For conservative force total Mechanical Energy at any point = E= P.E + K.E

∴ Total energy,

E = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2\)

∴ E = \(\frac{1}{2} m \omega^2\left\{A^2-x^2+x^2\right\}=\frac{1}{2} m \omega^2 A^2\)

So for a body in S.H.M total energy at any point of its motion is constant and equals to \(\frac{1}{2} m \omega^2 x^2\)

KSEEB Class 11 Physics Chapter 14 Oscillations Problems

Question 1. The bob of a pendulum is made of a hollow brass sphere. What happens to the time period of the pendulum, if the bob is filled with water completely? Why?
Solution:

If the hollow brass sphere is completely filled with water, then time period of the simple pendulum does not change. This is because time period of a pendulum is independent of the mass of the bob.

Question 2. Two identical springs of force constant “k” are joined one at the end of the other (in series). Find the effective force constant of the combination.
Solution:

When two springs of constant k each are joined together with end to end in series then effective spring constant \(\mathrm{k}=\frac{\mathrm{k}_1 \mathrm{k}_2}{\mathrm{k}_1+\mathrm{k}_2}\)

In this case \(k_{e q}=\frac{k \cdot k}{k+k}=\frac{k}{2}\)

In a series combination, the force constant of springs decreases.

Question 3. What are the physical quantities having maximum value at the mean position in SHM?
Solution:

In S.H.M at mean position velocity and kinetic energy will have maximum values.

KSEEB Class 11 Physics Solutions for Chapter 14 Oscillations

Question 4. A particle executes SHM such that, the maximum velocity during the oscillation is numerically equal to half the maximum acceleration. What is the time period?
Solution:

Given maximum velocity, \(\mathrm{V}_{\max }=\frac{1}{2}\) maximum acceleration \(\left(a_{\max }\right)\)

But \(V_{\max }=A \omega\) and \(\mathrm{a}_{\max }=\omega^2 \mathrm{~A}\)

∴ \(A \omega=\frac{1}{2} \cdot A \omega^2 \Rightarrow \omega=2\)

Time period of the body, \(\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}\)

= \(\pi \mathrm{sec}\)

Question 5. An amass of 2 kg attached to a spring of force constant 200 Nm^-1 makes 100 oscillations. What Is the time taken?
Solution:

Mass attached, m = 2 kg; Force constant, k = 260 N/m

∴ Time period of loaded spring, \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)

= \(2 \pi \sqrt{\frac{2}{260}}=0.5509 \mathrm{sec}\)

∴ Time for 100 oscillations = 100 x 0.551 = 55.1 sec

Question 6. A simple pendulum in a stationary lift has time period of T. What would be the effect on the time period when the lift

1. Moves up with uniform velocity
2. Moves down with uniform velocity
3. Moves up with uniform acceleration ‘a’
4. Moves down with uniform acceleration ‘a’
5. Begin to fall freely under gravity?

Solution:

1. When the lift moves up with uniform velocity i.e., a = 0, there would be no change in the time period of a simple pendulum.

2. When the lift moves down with uniform velocity i.e., a = 0, there would be no change in the time period of a simple pendulum.

3. When the lift is moving up with acceleration ‘a’ then relative acceleration = g + a

∴ Time period, \(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}+\mathrm{a}}}\) so when the lift is moving up with uniform acceleration time period of the pendulum in it decreases.

4. When lift is moving down with acceleration ‘a’ time period, \(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}-\mathrm{a}}}\)

(g – a = relative acceleration of pendulum)

So time period of the pendulum in the lift decreases:

5. If the lift falls freely, a = g then the time period of a simple pendulum becomes infinite.

Question 7. A particle executing SHM has an amplitude of 4cm, and its acceleration at a distance of 1cm from the mean position is 3 cm s^-2. What will its velocity be when it is at a distance of 2cm from its mean position?
Solution:

Amplitude, A = 4cm =4×10^-2 m

Acceleration, a = 3cm/s² = 3 x 10^-2 m/s²

Displacement, y = 1cm = 10^-2 m

∴ Angular velocity, \(\omega=\sqrt{\frac{a}{y}}=\sqrt{\frac{3}{1}}=\sqrt{3}\)

To find the velocity at a displacement of 2 cm use \(\mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{y}^2} \text { given } \mathrm{y}=2 \mathrm{~cm}\)

∴ V = \(\sqrt{3} \sqrt{(16-4) 10^{-4}}\)

= \(\sqrt{3} \times \sqrt{12} \cdot 10^{-2}=\sqrt{3} \times 2 \sqrt{3} \times 10^{-2}\)

= \(6 \times 10^{-2} \mathrm{~m} / \mathrm{sec} \approx 6 \mathrm{~cm} / \mathrm{sec}\)

Question 8. A simple harmonic oscillator has a time period of 2s. What will be the change in the phase after 0.25 s after leaving the mean position?
Solution:

Time period, T = 2 sec; time, t = 0.25 sec

Phase difference after t sec = \(\phi=\frac{t}{\mathrm{~T}} \times 2 \pi\)

= \(\frac{0.25}{2} \times 2 \pi=\frac{\pi}{4}=90^{\circ}\)

For a phase of \(\frac{\pi}{4}\) starting from the mean position the body will be at an extreme position. (Phase difference between mean position and extreme position is \(\frac{\pi}{4}\) Rad or 90°)

Question 9. A body describes simple harmonic motion with an amplitude of 5 cm and a period of  0.2 s. Find the acceleration and velocity of the body when the displacement is

1. 5 cm
2. 3 cm
3. 0 cm.

Solution:

Given that, A = 5 cm = 5 x 10^-2 m and T = 0.2 s

Angular velocity, ω = \(\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{0.2}=10 \pi \mathrm{rads}^{-1}\)

1. Displacement, \(y=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}\)

Acceleration of the body, a = \(-\omega^2 y\) = \(-(10 \pi)^2 \times 5 \times 10^{-2}=-5 \pi^2 \mathrm{~ms}^{-2}\)

Velocity of the body, v = \({ }_{10} \sqrt{\mathrm{A}^2-\mathrm{y}^2}\)

= \(10 \pi \sqrt{\left(5 \times 10^{-2}\right)^2-\left(5 \times 10^{-2}\right)^2}=0\)

2. Displacement, \(y=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}\)

Acceleration of the body, \(a=-\omega^2 y\)

= \(-(10 \pi)^2 \times 3 \times 10^{-2}=-3 \pi^2 \mathrm{~ms}^{-2}\)

Velocity of the body, \(v=\omega \sqrt{A^2-y^2}\)

= \(10 \pi \sqrt{\left(5 \times 10^{-2}\right)^2-\left(3 \times 10^{-2}\right)^2}\)

= \(10 \pi \times 4 \times 10^{-2}=0.4 \pi \mathrm{ms}^{-1}\)

3. Displacement, \(\mathrm{y}=0 \mathrm{~cm}\)

Acceleration of the body, \(a=-\omega^2 y=0\)

Velocity of the body, \(v=\omega \sqrt{\mathrm{A}^2-\mathrm{y}^2}\)

= \(10 \pi \sqrt{\left(5 \times 10^{-2}\right)^2-(0)^2}\)

= \(10 \pi \times 5 \times 10^{-2},=0.5 \pi \mathrm{ms}^{-1}\)

Question 10. The mass and radius of a planet are double that of the Earth. If the time period of a simple pendulum on the earth is T, find the time period on the planet.
Solution:

Mass of planet, \(\mathrm{M}_{\mathrm{p}}=2 \mathrm{M}_{\mathrm{e}}\);

The radius of the planet, \(R_p=2 R_e\)

The time period of the pendulum on earth =T;

Time period on planet \(=\mathrm{T}^{\prime}\)

∴ \(\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}}{\mathrm{R}^2} ; \mathrm{g}_{\mathrm{p}}=\frac{\mathrm{G} \cdot 2 \mathrm{M}}{(2 R)^2}=\frac{\mathrm{GM}}{2 \mathrm{R}^2}=\frac{\mathrm{g}_{\mathrm{e}}}{2}\)

T = \(2 \pi \sqrt{\frac{l}{\mathrm{~g}_e}}\), on the earth;- \(\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{l}{\mathrm{~g}_{\mathrm{p}}}}\), on the planet

∴ \(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}_e}{\mathrm{~g}_{\mathrm{p}}}}=\sqrt{2}\) or \(\mathrm{T}^{\prime}=\sqrt{2} \mathrm{~T}\)

Question 11. Calculate the change in the length of a simple pendulum of length 1, when its period of oscillation changes from 2 s to 1.5 s.
Solution:

For seconds pendulum \(T_1=2 \mathrm{sec}\);

Length \(l_1=1 \mathrm{~m}\).

New time period \(\mathrm{T}_2=1.5 \mathrm{sec}\); Length \(l_2=\)?

In Pendulum \(T=2 \pi \sqrt{\frac{l}{g}}\);

For second’s pendulum, \(2=2 \pi \sqrt{\frac{l}{g}}=2 \pi \sqrt{\frac{l}{g}}\)

On planet, \(1.5=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \Rightarrow \frac{2}{1.5}=\frac{1}{\sqrt{l}}=\frac{4}{3}\)

l = \(\frac{9}{16} \mathrm{~m}\)

∴ Decrease in length \(=1-\frac{9}{16}=\frac{7}{16} \mathrm{~m}\)

= \(0.4375 \mathrm{~m}\)

Question 12. A freely falling body takes 2 seconds to reach the ground on a plane when it is dropped from a height of 8m. If the period of a simple pendulum is seconds on the planet. Calculate the length of the pendulum.
Solution:

Height, \(\mathbf{h}=8 \mathrm{~m}\);

Time taken to reach the ground, \(\mathrm{t}=2 \mathrm{sec}\)

But for a body dropped, \(t=\sqrt{\frac{2 h}{g}}\)

⇒ 2 = \(\sqrt{\frac{16}{g}} \Rightarrow \mathrm{g}=\frac{16}{4}=4 \mathrm{~m} / \mathrm{s}^2\) on that planet Time period of pendulum, \(T=2 \pi \sqrt{\frac{l}{g}}=\pi\)

∴ \(2 \sqrt{\frac{l}{\mathrm{~g}}}=1\)

or \(\frac{l}{\mathrm{~g}}=\frac{1}{4} \Rightarrow l=\frac{\mathrm{g}}{4}\)

Length of pendulum \(=\frac{4}{4}=1 \mathrm{~m}=100 \mathrm{~cm}\) on that planet

KSEEB Class 11 Physics Chapter 14 Problems

Question 13. The period of a simple pendulum Is found to Increase by 50% when the length of the pendulum Is Increased by 0.6 m. Calculate the Initial length and the initial period of oscillation at a place where g = 9.8 m/s².
Solution:

1. Increase In length of pendulum = 0.6m;

Increase in time period = 50% = 1.5T

Let the original length of the pendulum = 1

Original time period = T; g = 9.8 m/s²

For 1st case \(9.8=4 \pi^2 \frac{l}{\mathrm{~T}^2} \rightarrow 1\);

For 2nd case \(l_1=(l+0.6), \mathrm{T}_1=.1 .5 \mathrm{~T}\)

9\(\times 8=4 \pi^2 \frac{l_1}{T_1^2} \rightarrow(2)\)

divide eq. (2) with eq. (1)

1 = \(\frac{l_1}{l} \cdot \frac{\mathrm{T}^2}{\mathrm{~T}_1^2} \Rightarrow \frac{l_1}{l}=\frac{\mathrm{T}_1^2}{\mathrm{~T}^2}=2.25 \Rightarrow l_1=2.25 l\)

But \(l_1=l+0.6\);

∴ \(l+0.6=2.25 l \Rightarrow 0.6=1.25 l\)

∴ Length of pendulum \(l=\frac{0.6}{1.25}=0.48 \mathrm{~m}\)

2. Time period \(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\)

= \(2 \times 3.142 \sqrt{\frac{0.48}{9.8}}\)

= \(6.284 \times 0.2213\)

= \(1.391 \mathrm{sec}\)

Question 14. A clock regulated by a second pendulum keeps the correct time. During summer the length of the pendulum increases to 1.02m. How much will the clock gain or lose in one day?
Solution:

The time period of the seconds’ pendulum, T = 2 sec

Length of seconds pendulum, \(\mathrm{L}=\mathrm{gT}^2 / 4 \pi^2=0.9927 \mathrm{~m}\)

Length of seconds pendulum during summer \(=1.02 \mathrm{~m}\)

∴ Error in length, \(\Delta l=1.02-1=0.0273\)

In pendulum \(\mathrm{T} \propto \sqrt{l}\).

From principles of error \(\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta l}{l}=\frac{1}{2} \frac{0.0273}{0.9927}\)

∴ Error in time per day

= \(86,400 \times \frac{1}{2} \frac{0.0273}{0.9927}=1188 \mathrm{sec} \text {. }\)

Oscillations questions and answers KSEEB Physics

Question 15. The time period of a body suspended from a spring is T. What will be the new time period if the spring is cut into two equal parts and

1. The mass is suspended from one part?
2. The mass is suspended simultaneously from both the parts?

Solution:

Time period of spring, \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)

When a spring is cut into two equal parts force constant of each part K = 2K

1. For one piece, \(T=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_1}}\)

= \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}} \frac{1}{\sqrt{2}}=\frac{\mathrm{T}}{\sqrt{2}}\)

2. When mass is suspended simultaneously from two parts ⇒ they are connected in parallel. For springs in parallel \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_1+\mathrm{K}_2=4 \mathrm{~K}\)

Time period, \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_{\mathrm{p}}}}\)

= \(2 \pi \sqrt{\frac{\mathrm{m}}{4 \cdot \mathrm{K}}}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}} \cdot \frac{1}{2}=\frac{\mathrm{T}}{2}\)

Question 16. What is the length of a second pendulum on the earth?
Solution:

g = \(4 \pi^2 \frac{l}{\mathrm{~T}^2}, l=\frac{\mathrm{gT}^2}{4 \pi^2}\)

= \(\frac{9.8 \times 2^2}{4 \times(3.14)^2}=\frac{9.8}{9.86}\)

= \(0.99=1 \mathrm{~m}=100 \mathrm{~cm}\)

KSEEB Solutions For Class 11 Physics Chapter 7 Systems Of Particles And Rotational Motion Important Points

Rigid Body: A rigid body is a body with a perfectly definite and unchanging shape. The distance between all the pairs of particles of such a body does not change.

There is no real body that is truly rigid. All real bodies deform under the influence of forces., But in many cases this deformation is negligible.

Translational Motion: In translational motion, the body will move as a whole from one place to another place.

In pure translational motion, all the particles of the body will have the same velocity at any instant in time.

Axis Of Rotation: To prevent translational motion a rigid body has to be fixed along a straight line. Then the only possible motion is rotation about that fixed line. This fixed line is called the axis of rotation.

Read and Learn More KSEEB Class 11 Physics Solutions

Rotation: In rotation, every particle of a rigid body moves in a circle which lies in a plane perpendicular to the axis of rotation. The rotating body has a centre on the axis.

Centre Of Mass: For a rigid body or system of particles the total mass seems to be concentrated at a particular point is called the centre of mass. Such a point will behave as if it represents the whole translational motion of that body.

Centre Of Gravity: It is a point in the body where the body’s total weight seems to be concentrated.

If we apply an equal and opposite force to the body’s weight (W = mg) the body will be in mechanical equilibrium (i.e. both in translational and rotational equilibrium).

Co-ordinates Of The Centre Of Mass:

1. For symmetric bodies when the origin is taken at the geometric centre then the centre of mass is also at the geometric centre. Example: Thin rod, disc, sphere etc.
2. Let two bodies of masses say m₁ and m₂ are at distances say x₁ and x₂ from origin then centre of mass \(x_c=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

So we can assume that the position centre of mass is the ratio of the sum of the moment of masses and the total mass of the body.

1. Let two bodies of equal masses m and I m are separated by a distance x then the centre of mass of that system is at \(\frac{x}{2}\).
2. If three equal masses are at the corners of a triangle then centre of mass is at  centroid of that triangle.
3. Let a system of particles say m₁, m₂, m₃ …. m_n are in a plane or in space then co-ordinates of centre of mass will have x, y for the plane and X, Y and Z for space where

⇒ \(X_c =\frac{m_1 x_1+m_2 x_2+\ldots \ldots \ldots .+m_n x_n}{m_1+m_2+\ldots \ldots \ldots .+m_n}\)

⇒ \(Y_c=\frac{m_1 y_1+m_2 y_2+\ldots \ldots \ldots . .+m_n y_n}{m_1+m_2+\ldots \ldots \ldots . .+m_n}\)

⇒ \(Z_c=\frac{m_1 z_1+m_2 z_2+\ldots \ldots \ldots .+m_n z_n}{m_1+m_2+\ldots \ldots \ldots .+m_n}\)

Characteristics Of Centre Of Mass:

1. The total mass of the body seems to be concentrated at the centre of mass.
2. The total external force (F_ext) applied on a body seems to be applied at the centre of mass. F_ext = M a_c where ‘a_c‘ is the acceleration of the centre of mass.
3. Internal forces can’t change the motion of center of mass.
4. A complex motion Is a combination of translational and rotational motions. In complex motion centre of mass represents the entire translational motion of the whole body.
5. The momentum of a body is the product of the mass of the body and velocity of the centre of mass \(\overline{\mathrm{P}}=\mathrm{MV}_{\mathrm{c}}\)
6. The coordinates of the centre of mass do not depend on the coordinate system chosen.

Motion Of Centre Of Mass:

1. The motion of the centre of mass represents the translational motion of the whole body.

2. Velocity of centre of mass \(\mathrm{V}_{\mathrm{c}}=\frac{\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2+\ldots \ldots \ldots+\mathrm{m}_{\mathrm{n}} \mathrm{v}_{\mathrm{n}}}{\Sigma \mathrm{m}_{\mathrm{i}}}\)

i.e., the velocity of the centre of mass is the ratio of the sum of the momentum of all particles to the total mass of the body.

3. The momentum of the centre of mass \(\overline{\mathrm{P}}_{\mathrm{c}}\) is the sum of the momentum of all the particles of the body.

⇒ \(\bar{P}_c=M \bar{V}_c=m_1 v_1+m_2 v_2+\ldots \ldots \ldots . .+m_n v_n\)

4. The external force acting on the centre of mass \(F_{e t t}=M A_c=m_1 a_1+m_2 a_2 \ldots \ldots+m_n a_n\) or \(F_{e x t}=M A_c=F_1+F_2+\ldots \ldots. .+F_n\)

Where a₁, a₂, a_n are accelerations of individual particles of masses m₁, m₂…. m_n

Explosion Of A Shell In Mid-Air: Let a shell move along a parabolic trajectory explode in mid-air and divide into a number of fragments. Still, the centre of mass of that system of fragments will follow “the same parabolic path”.

Explanation: Explosion is due to internal forces. Internal forces cannot change the momentum of a body. So algebraic sum of the momentum of all fragments is constant. So the velocity of the centre of mass is constant. Hence centre of mass will follow the same parabolic path.

Crow Product Or Vector Product Of Vectors: If the multiplication of two vectors generates a vector then that vector multiplication is called cross product. Mathematically \(\bar{A} \times \bar{B}=|\bar{A}||\bar{B}| \sin \theta \hat{n}\)

Where \(\hat{n}\) is a unit vector perpendicular to the plane of \(\bar{A}\) and \(\bar{B}\).

The new vector generated is always perpendicular to both \(\bar{A}\) and \(\bar{B}\) i.e., perpendicular to the plane containing \(\bar{A}\) and\(\bar{B}\)

Systems of Particles and Rotational Motion KSEEB Physics Chapter 7 

Properties Of Cross Product:

1. Cross product is not commutative i.e., \(\overline{\mathrm{A}} \times \overline{\mathrm{B}} \neq \overline{\mathrm{B}} \times \overline{\mathrm{A}} \text { But } \overline{\mathrm{A}} \times \overline{\mathrm{B}}=-\overline{\mathrm{B}} \times \overline{\mathrm{A}}\)
2. Cross product obeys distributive law i.e. \(\overline{\mathrm{A}} \times(\overline{\mathrm{B}}+\overline{\mathrm{C}})=(\overline{\overline{\mathrm{A}}} \times \overline{\mathrm{B}})+(\overline{\mathrm{A}} \times \overline{\mathrm{C}})\)
3. If any vector is represented by the combination of \(\bar{i}\), \(\bar{j}\) and \(\bar{k}\) then cross product will obey right-hand screw rule.
4. The product of two coplanar perpendicular unit vectors will generate a unit vector perpendicular to that plane i.e. \(\overline{\mathrm{i}} \times \overline{\mathrm{j}}=\overline{\mathrm{k}}\), \(\overline{\mathrm{j}} \times \overline{\mathrm{k}}=\overline{\mathrm{i}}\) and \(\cdot \overline{\mathrm{k}} \times \overline{\mathrm{i}}=\overline{\mathrm{j}}\)
5. Cross product of parallel vectors is zero i.e. \(\overline{\mathrm{i}} \times \overline{\mathrm{i}}=\overline{\mathrm{j}} \times \overline{\mathrm{j}}=\overline{\mathrm{k}} \times \overline{\mathrm{k}}=0\)

Angular Displacement (θ): The angle subtended by a body at the centre when it is in angular motion is called angular displacement. Unit: Radian.

Angular Velocity (ω): The rate of change in angular displacement is called angular velocity.

Let angular displacement be Δθ over a  time interval Δt then average angular velocity \(\omega=\frac{\Delta \theta}{\Delta \mathrm{t}}\), Unit = Radian/sec

1. The relation between angular velocity and| linear velocity is v = rω
2. For a body in rotatory motion, all the particles will have the same angular velocity ‘ω’. But linear velocity ‘v’ changes.

Angular Acceleration(α): The rate of change in angular velocity is defined as angular acceleration.

Angular acceleration \(\alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}\).

Unit: Radian /sec²

Moment Of Force Couple Or Torque (τ): The moment of force is called torque or moment of force couple.

Let a force \(\overrightarrow{\mathrm{F}}\) be applied on a point ‘P’.

The position vector of P from the origin is \(\overrightarrow{\mathrm{r}}\) then

Torque \(\tau=\overline{\mathrm{r}} \times \overline{\mathrm{F}}=|\overline{\mathrm{r}}| \times|\overline{\mathrm{F}}| \sin \theta \cdot \overrightarrow{\mathrm{n}}\)

It is a vector. Its direction is perpendicular to both \(\overrightarrow{\mathrm{r}}\) and \(\overrightarrow{\mathrm{F}}\) (or) it is perpendicular to the plane containing \(\overrightarrow{\mathrm{r}}\) and \(\overrightarrow{\mathrm{F}}\).

Unit: Newton metre (N – m); D.F: ML² T^-2

1. Torque is the product of force \(\vec{F}\) and the perpendicular distance between force (F) and point of application (i.e. r sin θ).
2. Torque represents the energy with which a body is rotated. So units of torque and energy are the same.

Angular Momentum (L): Let a particle of mass ‘m’ has a linear momentum \(\overrightarrow{\mathrm{p}}\) and its position vector is \(\overrightarrow{\mathrm{r}}\) from origin then angular momentum of that particle is defined as \(\overline{\mathrm{L}}=\overline{\mathrm{r}} \times \overline{\mathrm{p}}=|\overline{\mathrm{r}} \| \overline{\mathrm{p}}| \sin \theta \cdot \hat{\mathrm{n}}\)

Angular momentum is a vector. \(\overrightarrow{\mathrm{L}}\) is perpendicular to the plane containing \(\overrightarrow{\mathrm{r}}\) and \(\overrightarrow{\mathrm{p}}\)

Unit: Kg – metre² and D.F.: ML²

Angular momentum is the product of momentum \(\bar{P}\) and perpendicular distance (r sin θ) from the origin. It is also written as L = Iω

Torque And Angular Momentum: The time rate of change of the angular momentum of a particle is equal to the torque acting on it.

Torque \(\tau=\frac{\mathrm{d} \overline{\mathrm{L}}}{\mathrm{dt}}=\overline{\mathrm{r}} \cdot \frac{\mathrm{d} \overline{\mathrm{p}}}{\mathrm{dt}}\)

The time rate of change of the angular momentum of a system of particles about a point is equal to the sum of external torque i.e., \(\frac{\mathrm{d} \overline{\mathrm{L}}}{\mathrm{dt}}=\tau_{\text {ext }}\) just like \(\frac{\mathrm{d} \overline{\mathrm{p}}}{\mathrm{dt}}=\mathrm{F}_{\text {ext }}\)

Law Of Conservation Of Angular Momentum: When external torque (τ_ext) is zero, the total angular momentum of a system is conserved i.e., it remains constant.

When \(\tau_{\text {ext }}=0\) then \(\frac{d \overline{\mathrm{L}}}{d t}=0\) i.e. angular momentum \(\bar{L}\) is constant, i.e. \(\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2=\) constant

1. For a system of particles when \(\tau_{\text {ext }}=0\) then \(\mathrm{d} \overline{\mathrm{L}}_1+\mathrm{d} \overline{\mathrm{L}}_2+\ldots \ldots+\mathrm{d} \overline{\mathrm{L}}_{\mathrm{n}}=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{d} \overline{\mathrm{L}}_{\mathrm{i}}=0\)
2. The law of conservation of angular momentum is similar to the law of conservation of linear momentum in linear motion.

Equilibrium Of A Rigid Body: A rigid body is said to be in mechanical equilibrium if both its linear momentum and angular momentum are not changing with time. Then the body has neither linear acceleration nor angular acceleration.

(or),

1. The vector sum of all the forces acting on a rigid body must be zero.

i.e. \(F_1+F_2+\ldots \ldots \ldots+F_n=\sum_{i=1}^n F_i=0 \text {. }\)

This condition provides translational equilibrium of the body.

The vector sum of all the torques acting on a rigid body must be zero i.e., \(\tau_1+\tau_2 \ldots \ldots \ldots .+\tau_n=\sum_{i=1}^n \tau_i=0\)

This condition provides the rotational equilibrium of the body.

Couple (Or) Force Couple: A pair of equal and opposite forces with different lines of action is known as a couple. A couple produces rotation without translation.

Systems Of Particles and Rotational Motion Questions And Answers KSEEB Physics

Moments Or Moment Of Force: It is defined as the product of force and the perpendicular distance between the force and its point of application

Principles Of Moments: For a lever to be in mechanical equilibrium let R be the reaction of the support at the fulcrum. It is directed upwards and F₁ and F₂ are the forces then

For translational equilibrium R – F₁ – F₂ = 0 i.e. algebraic sum of forces must be zero.

For rotational equilibrium d₁F₁ – d₂F₂ = 0. i.e. algebraic sum of moments must be zero.

Lever: An ideal lever is a light rod pivoted at a point along its length. This point is called “fulcrum”. In levers d₁F₁ = d₂F₂ i.e. Load arm x load = Force arm x Force

Mechanical advantage (M.A) = \(\frac{F_1}{F_2}=\frac{d_2}{d_2}\)

If effort arm d₂ is larger than the load arm then M.A > 1 i.e. we can lift greater loads with less effort.

Moment Of Inertia(I): The inertia of a rotating body is called moment of inertia.

Mathematically, Moment of Inertia, \(I=\sum_{i=1}^n m_1 r_1^2=M R^2\)

Radius Of Gyration (k): The radium of gyration of a body about an axis may be defined as the distance from the axis of a mass point whose mass is equal to the whole mass of the body and whose moment of inertia is equal to the moment of inertia of the whole body about that axis.

Fly Wheel: Flywheel is a metallic body with large moments of inertia.

It is used in the rotational motion of engines like automobiles. It allows a gradual change in speed and prevents jerky motion.

Perpendicular Axis Theorem: The moment of inertia of a plane body (lamina) about an axis perpendicular to its plane is equal to the sum of its moment of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body, i.e., l_z =I_x+ I_y

Parallel Axis Theorem: The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through the centre of mass and the product of its mass and the square of the distance between the two parallel axes, i.e., I = I_G + MR²

Rolling Motion: Rolling motion is a combination of translational motion and mandatory motion.

Kinetic Energy Of A Rolling Body: When a body is rolling on a body without slipping then it will have translational kinetic energy \(\left(\frac{1}{2} m v^2\right)\) and rotational kinetic energy \(\left(\frac{1}{2} I \omega^2\right)\).

∴ Total kinetic energy of rolling body \(\mathrm{K} \cdot \mathrm{E}_{\mathrm{R}}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\)

Systems Of Particles And Rotational Motion Important Formulae

For a two-particle system of masses m₁ and m₂ with positions x₁ and x₂.

1. Coordinates of centre of mass, \(x_c=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)
2. If coordinate system coincides with m then \(x_c=\frac{m_2 x_2}{m_1+m_2} \text { or } x_c=\frac{m_2 d}{m_1+m_2}\) where d is distance between m₁ and m₂.
3. Ratio of distances from centre of mass is \(\frac{d_1}{d_2}=\frac{m_2}{m_1}\)

For Many-Particle System

1. \(x_{c . m}=\frac{m_1 x_1+m_2 x_2+m_3 x_3+\ldots .+m_n x_n}{m_1+m_2+m_3+\ldots .+m_n}\)

= \(\frac{1}{M} \sum_{i=1}^n m_1 x_1\)

2. \(Y_{c. m}=\frac{m_1 y_1+m_2 y_2+m_3 y_3+\ldots+m_n y_n}{m_1+m_2+m_3+\ldots .+m_n}\)

= \(\frac{1}{\mathrm{M}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{m}_{\mathrm{i}} \mathrm{y}_{\mathrm{i}}\)

= \(\frac{1}{M} \sum_{i=1}^n \mathrm{~m}_i z_i\)

4. Position vector of the centre of mass \(\overline{\mathrm{r}}_{\mathrm{C}. \mathrm{M}}=\frac{\mathrm{m}_1 \overline{\mathrm{r}}_1+\mathrm{m}_2 \overline{\mathrm{r}}_2+\mathrm{m}_3 \overline{\mathrm{r}}_3+\ldots+m_n \overline{\mathrm{r}}_{\mathrm{n}}}{m_1+m_2+m_3+\ldots+m_n}\)

= \(\frac{1}{M} \sum_{\mathrm{i}=1}^n m_i r_i\)

5. Momentum of centre of mass, \(P_c=m v_c\)

= \(\sum_{i=1}^n \mathrm{~m}_1 v_i \text { or } \overline{\mathrm{P}}_{\mathrm{C}}=\overline{\mathrm{P}}_1+\overline{\mathrm{P}}_2+\ldots . . \overline{\mathrm{P}}_{\mathrm{n}}\) or \(\mathrm{m}_1 \overline{\mathrm{v}}_1+\mathrm{m}_2 \overline{\mathrm{v}}_2+\ldots . .+\mathrm{m}_{\mathrm{n}} \overline{\mathrm{v}}_{\mathrm{n}}\)

The velocity of the centre of mass \(V_{c m}=\frac{m_1 v_1+m_2 v_2+\ldots .+m_n v_n}{m_1+m_2+\ldots. .+m_n}\)

6. Acceleration of centre of mass, a = \(\frac{F_C}{M}=\sum_{i=1}^n a_i m_1\)

∴ a = \(\frac{m_1 a_1+m_2 a_2+m_3 a_3+\ldots .+m_n a_n}{m_1+m_2+m_3+\ldots .+m_n}\)

Cross Product: \(\bar{A}\) x \(\bar{B}\) is defined as |\(\bar{A}\)| |\(\bar{A}\)| sin G. \(\bar{n}\) where \(\bar{n}\) is a unit vector perpendicular to the plane of \(\bar{A}\) and \(\bar{B}\).

In cross product \(\overline{\mathrm{i}} \times \overline{\mathrm{i}}=\overline{\mathrm{j}} \times \overline{\mathrm{j}}=\overline{\mathrm{k}} \times \overline{\mathrm{k}}=0\) i.e., cross product of parallel vectors is zero.

In cross product \(\overline{\mathrm{i}} \times \overline{\mathrm{j}}=\overline{\mathrm{k}}, \overline{\mathrm{j}} \times \overline{\mathrm{k}}=\overline{\mathrm{i}}\) and \(\overline{\mathrm{k}} \times \overline{\mathrm{i}}=\overline{\mathrm{j}}\) cross product of two heterogeneous unit vectors will generate third unit vector) taken in a clockwise direction.

If \(\bar{A}=x_1 \bar{i}+y_1 \bar{j}+z_1 \bar{k}\) and \(\bar{B}=x_2 \bar{i}+y_2 \bar{j}\) + \(z_2 \bar{k}\) then

⇒ \(\bar{A} \times \bar{B}\) = \(\left|\begin{array}{rrr}
i & j & k \\
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2
\end{array}\right|\)

= \(\left(y_1 z_2-y_2 z_1\right) \bar{i}-\left(x_1 z_2-x_2 z_1\right) \bar{j}+\left(x_1 y_2-x_2 y_1\right) \bar{k}\)

Angular velocity, \(\omega=\frac{\text { angular displacement }}{\text { time }}=\frac{\theta}{t}\)

∴ \(\omega=\frac{\theta}{t}\)

For small quantities, \(\omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}\);

Unit: Radian/sec.

∴ \(\omega=\frac{\theta}{\mathrm{t}}\) or \(\omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}\) or \(\omega=\frac{2 \pi \mathrm{n}}{\mathrm{t}}(\mathrm{n}=\) number of rotations)

Angular acceleration, \(\alpha=\frac{\text { change in angular velocity }}{\text { time }}\)

∴ \(\alpha=\frac{\omega_2-\omega_1}{\mathrm{t}}\) or

∴ \(\alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}\) Unit : Radian \(\mathrm{sec}^2\)

The relation between v and ro is v = rω

The relation between a and α is a = rα

Centripetal acceleration, \(a_c=r w^2=v \omega=\frac{v^2}{r}\)

Centrifugal force = \(\frac{m v^2}{r}=m r \omega^2\)

When a coin is placed on a gramophone disc or on a circular turn table or for a vehicle is moving in a curved path limiting friction, \(\mu_s=\frac{f_s}{N}=\frac{r \omega^2}{g}\)

When a vertically hanging body ‘M’ is balanced by a rotating body m in the horizontal plane then at equilibrium

Mg = mrω² or Angular velocity required for balance is \(\omega=\sqrt{\frac{\mathrm{Mg}}{\mathrm{mr}}}\)

Torque, \(\tau=\overline{\mathrm{r}} \times \overline{\mathrm{F}}=|\overline{\mathrm{r}}||\overline{\mathrm{F}}| \sin \theta\). It represents the energy with which a body is turned.

Moment of force couple = one of the forces in couple x distance between the directions of forces.

Moment of inertia, \(I=\sum_{i=1}^n m_i r_1^2 \text { or } I=M R^2\)

If the moment of inertia, I = MR² = MK² then K is called radius of gyration.

From the parallel axis theorem moment of inertia about any parallel axis to the axis passing through the centre of mass is, I = I_G + MR².

From perpendicular axis theorem M.O.I. about a perpendicular axis to the plane I_z = I_x+ I_y

KSEEB Class 11 Physics Chapter 7 Systems of Particles and Rotational Motion

Moment Of Inertia Of A Thin Rod Of Length l

1. M.O.I of the thin rod about its axis and perpendicular to the length, \(\mathrm{I}=\frac{\mathrm{M} l^2}{12}; \mathrm{K}=\frac{l}{\sqrt{12}}\)

2. M.O.I of the thin rod about the end of the rod and perpendicular to the length.

Moment Of Inertia Of A Circular Ring Of Radius R

1. M.O.I of a circular ring about an axis passing through the centre and perpendicular to its plane I = MR², K = R

2. M.O.I. of a circular ring about any diameter, \(\mathrm{I}=\frac{\mathrm{MR}^2}{2}; \mathrm{K}=\frac{\mathrm{R}}{\sqrt{2}}\)

3. M.O.I. about any tangent and parallel to the diameter \(\mathrm{I}=\frac{3}{2} \mathrm{MR}^2; \mathrm{K}=\sqrt{\frac{3}{2}} \mathrm{R}\)

Moment Of Inertia Of A Disc Of Radius R

1. M.O.I about an axis passing through the centre and perpendicular to the plane, \(\mathrm{I}=\frac{\mathrm{MR}^2}{2}; \mathrm{K}=\frac{\mathrm{R}}{\sqrt{2}}\)

2.  M.O.I. about any diameter, \(\mathrm{I}=\frac{\mathrm{MR}^2}{4} ; \mathrm{K}=\frac{\mathrm{R}}{2}\)

3. M.O.I. about any tangent \(\mathrm{I}=\frac{5}{4} \mathrm{MR}^2 ; \mathrm{K}=\frac{\sqrt{5}}{2} \mathrm{R}\)

Moment Of Inertia Of A Rectangular Plane Lamina

1. About the centre and perpendicular to the plane \(\mathrm{I}=\mathrm{M} \frac{\left(l^2+\mathrm{b}^2\right)}{12}; \mathrm{K}=\sqrt{\frac{l^2+\mathrm{b}^2}{12}}\)

2. About the axis parallel to length, \(\mathrm{I}=\frac{\mathrm{Mb}^2}{12}; K=\frac{\mathrm{b}}{\sqrt{12}}\)

3. About any axis parallel to breadth, \(\mathrm{I}=\frac{\mathrm{M} l^2}{12}; \mathrm{K}=\frac{l}{\sqrt{12}}\)

KSEEB Class 11 Physics Solutions For Chapter 7 Systems of Particles and Rotational Motion

Moment Of Inertia Of A Solid Sphere

1. About an axis passing through diameter \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^2; \mathrm{K}=\sqrt{\frac{2}{5}} \mathrm{R}\)

2. About any tangent, \(I=\frac{7}{5} M R^2 ; K=\sqrt{\frac{7}{5}} R\)

Moment Of Inertia Of A Hollow Sphere

1. About any diameter, \(\mathrm{I}=\frac{2}{3} \mathrm{MR}^2 ; \mathrm{K}=\sqrt{\frac{2}{3}} \mathrm{R}\)

2. About any tangent, \(\mathrm{I}=\frac{5}{3} \mathrm{MR}^2 ; \mathrm{K}=\sqrt{\frac{5}{3}} \mathrm{R}\)

Moment Of Inertia Of A Solid Cylinder Of Length L And Radius R

1. M.O.I. about natural axis of cylinder, \(\mathrm{I}=\frac{\mathrm{MR}^2}{2} ; \mathrm{K}=\frac{\mathrm{R}}{\sqrt{2}}\)

2. M.O.I. about an axis perpendicular to length and passing through centre, \(\mathrm{I}=\mathrm{M}\left(\frac{l^2}{12}+\frac{\mathrm{R}^2}{4}\right); \mathrm{K}=\sqrt{\frac{l^2}{12}+\frac{\mathrm{R}^2}{4}}\)

M.O.I Of A Hollow Cylinder Ol Length L And Radius R

1. About The natural axis, I = MR²; K = R

2. About an axis perpendicular to length and passing through centre \(\mathrm{I}=\mathrm{M}\left(\frac{l^2}{12}+\frac{\mathrm{R}^2}{2}\right); \mathrm{K}=\sqrt{\frac{l^2}{12}+\frac{\mathrm{R}^2}{2}}\)

Angular momentum, L = Iω

Relation between angular momentum (L) and torque \((\tau)\) is, \(\tau=\frac{d \overline{\mathrm{L}}}{d \mathrm{It}}=\frac{\mathrm{L}_2-\mathrm{L}_1}{1}\)

Relation between τ and α Is τ = lα

From the law of conservation of angular momentum I₁ω₁ + I₂ω₂ = constant (When no external torque on the body)

Systems Of Particles And Rotational Motion Solutions KSEEB Class 11 Physics Very Short Answer Questions

Question 1. Is it necessary that a mass should be pre¬sent at the centre of mass of any system?
Answer:

No. It is not necessary to present some mass at the centre of mass of the system.

Example: At the centre of the ring (or) bangle, there is no mass present at the centre of mass.

Question 2. What is the difference in the positions of a girl carrying a hag in one of her hands and another girl carrying a hag in each of her two hands?
Answer:

1. When she carries a bag in one hand her centre of mass will shift to the side of the hand that carries the bag.
2. When a bag is in one hand some un-balanced force will act on her and it is difficult to carry.
3. If she carries two bags in two hands then her centre of mass remains unchanged Force on the two hands is equal, and balanced so it Is easy to carry the bags.

Question 3. Two rigid bodies have the same moment of inertia about their axes of symmetry. Of the two, which body will have greater kinetic energy?
Answer:

The relation between angular momentum and kinetic energy is, \(K E = \frac{L^2}{21}\)

Because the moment, of inertia, is the same the body with large angular momentum will have larger kinetic energy.

Question 4. Why are spokes provided in a bicycle wheel?
Answer:

The spokes of the cycle wheel increase its moment of inertia. The greater the moment of inertia, the greater the opposition to any change in uniform rotational motion

. As a result, the cycle runs smoother and speeder. If the cycle wheel had no spokes, the cycle would be driven in jerks and hence unsafe.

Systems of Particles and Rotational Motion solutions KSEEB Class 11 Physics

Question 5. We cannot open or close the door by applying force to the hinges. Why?
Answer:

To open or close a door, we apply a force normal to the door. If the force is applied at the hinges the perpendicular distance of force is zero. Hence, there will be no turning effect however large force is applied.

Question 6. Why do we prefer a spanner of the longer arm as compared to the spanner of the shorter arm?
Answer:

The turning effect of force, \(\tau=\vec{r} \times \vec{F}\). When the arm of the spanner is long, r is larger. Therefore, a smaller force (F) will produce the same turning effect. Hence, the spanner of a longer arm is preferred as compared to the spanner of a shorter arm.

Question 7. By spinning eggs on a tabletop, how will you distinguish a hard-boiled egg from a raw egg?
Answer:

To distinguish between a hard-boiled egg and a raw egg, we spin each on a tabletop. The egg which spins at a slower rate shall be a raw egg. This is because, in a raw egg, liquid matter inside tries to get away from the axis of rotation.

Therefore, its moment of inertia ‘I’ increases. As τ = lα = constant, therefore, α decreased i.e., raw egg will spin With smaller angular acceleration.

Question 8. Why should a helicopter necessarily have two propellers?
Answer:

If the helicopter had only one propeller, then due to the conservation of angular momentum, the helicopter itself would turn in the opposite direction. Hence, the helicopter should necessarily have two propellers.

Question 9. If the polar ice caps of the earth were to melt, what would the effect of the length of the day be?
Answer:

Earth rotates about its polar axis. When the ice of the polar caps of the earth melts, mass concentrated near the axis of rotation spreads out. Therefore, a moment of inertia ‘I’ increases.

As no external torque acts, \(\mathrm{L}=\mathrm{I} \omega=\mathrm{I}\left(\frac{2 \pi}{\mathrm{T}}\right)\) = constant with the increase of I, T will increase i.e., the length of the day will increase.

Question 10. Why is it easier to balance a bicycle in motion?
Answer:

A bicycle in motion is in rotational equilibrium. From the principles of the Dynamics of rotational bodies is that the forces that are perpendicular to the axis of rotation will try to turn the axis of rotation. Still, necessary forces will arise cancelling these forces due to the inertia of rotation and the fixed position of the axis is maintained. So it is easy to balance a rotating body.

KSEEB Class 11 Physics Chapter 7 Systems Of Particles And Rotational Motion Short Answer Questions

Question 1. Distinguish between the centre of mass and the centre of gravity.
Answer:

Question 2. Show that a system of particles moving under the influence of external force moves as if the force is applied at a centre of mass.
Answer:

Consider a system of particles of masses \(\mathrm{m}_1, \mathrm{~m}_2, \ldots \ldots \mathrm{m}_{\mathrm{n}}\) moves with velocity

vectors \(\overrightarrow{v_1}, \overrightarrow{v_2}, \overrightarrow{v_3} \ldots ., \overline{v_n}\)

∴ Velocity of the centre of mass

⇒ \(\overrightarrow{\mathrm{V}}_{\mathrm{CM}}=\frac{1}{M}\left\{m_1 \overline{\mathrm{v}_1}+m_2 \overrightarrow{\mathrm{v}_2}+\ldots+m_n \overline{v_n}\right\}\)

∴ \(m_1+m_2+\ldots \ldots+m_n=M\)

We know

∴ \(\overline{\mathrm{a}}=\frac{\mathrm{d}}{\mathrm{dt}}(\overline{\mathrm{V}}) \text {. }\)

∴ Acceleration of centre of mass  \(\mathrm{a}_{\mathrm{CM}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{\mathrm{v}}_{\mathrm{CM}}\right)\)

∴ \(\mathrm{a}_{\mathrm{CM}}=\frac{1}{\mathrm{M}}\left\{\mathrm{m}_1 \frac{\mathrm{d}}{\mathrm{dt}} \vec{v}_1+\mathrm{m}_2 \frac{\mathrm{d}}{\mathrm{dt}} \overrightarrow{\mathrm{v}_2}+\ldots .+\mathrm{m}_{\mathrm{n}} \frac{\mathrm{d}}{\mathrm{dt}} \overrightarrow{\mathrm{v}_{\mathrm{n}}}\right\}\)

But Force (F)=m a, so the total force on the body is

⇒ \(\mathrm{F}=\mathrm{Ma}_{\mathrm{C} . \mathrm{M}}=\mathrm{m}_1 \mathrm{a}_1+\mathrm{m}_2 \mathrm{a}_2+\mathrm{m}_3 \mathrm{a}_3+\ldots \ldots+\mathrm{m}_{\mathrm{n}} \mathrm{a}_{\mathrm{n}}\)

or Total Force \(\mathrm{F}=\mathrm{Ma}_{\mathrm{C} . \mathrm{M}}\).

= \(\mathrm{F}_1+\mathrm{F}_2+\mathrm{F}_3\) + \(\ldots \ldots . .+\mathrm{F}_{\mathrm{n}}\)

Hence, the total force on the body is the sum of forces on individual particles and it is equal to the force on the centre of mass of the body.

KSEEB Class 11 Physics Chapter 7 Systems Of Particles And Rotational Motion

Question 3. Explain the centre of mass of the earth-moon system and its rotation around the sun.
Answer:

The interaction of the earth and moon does not affect the motion of the centre of mass of the earth and moon system around the sun.

• The gravitational force between the earth and the moon is the internal force. Internal forces cannot change the position of the centre of mass.
• The external force acting on the centre of mass of the Earth and Moon system is the force between the sun and C.M. of the Earth and Moon system.
• The motion of the centre of mass depends on external force. Hence, the moon system continues to move in an elliptical path around the sun. It is irrespective of the rotation of the moon around Earth.

Question 4. Define vector product. Explain the properties of a vector product with two examples.
Answer:

Vector Product (Or) Cross Product: If the product of two vectors (say \(\bar{A}\) and \(\bar{B}\)) gives a vector then that multiplication of vectors is called cross product or vector product of vectors.

Mathematically \(\bar{A}\) x \(\bar{A}\) = |\(\bar{A}\)| |\(\bar{B}\)|sinθ-\(\hat{n}\)

where \(\hat{n}\) = a unit vector perpendicular to both \(\bar{A}\) and \(\bar{B}\) (or) perpendicular to the plane of \(\bar{A}\) and \(\bar{B}\).

Properties Of Cross Product:

1. Cross product is not commutative i.e. \(\bar{A} \times \bar{B} \neq \bar{B} \times \bar{A} \text { But } \bar{A} \times \bar{B}=-\bar{B} \times \bar{A}\)
2. Cross product obeys distributive law i.e., \(\bar{A} \times(\bar{B}+\bar{C})=(\bar{A} \times \bar{B})+(\bar{A} \times \bar{C})\)
3. If any vector is represented by the combination of \(\bar{i}\), \(\bar{j}\) and \(\bar{k}\) then cross product will obey right-hand screw rule.
4. The product of two coplanar perpendicular unit vectors will generate a unit, vector perpendicular to that plane i.e., \(\overline{\mathrm{i}} \times \overline{\mathrm{j}}=\overline{\mathrm{k}}, \overline{\mathrm{j}} \times \overline{\mathrm{k}}=\overline{\mathrm{i}}\) and \(\overline{\mathrm{k}} \times \overline{\mathrm{i}}=\overline{\mathrm{j}}\)
5. The cross product of parallel vectors is zero i.e., \(\overline{\mathrm{i}} \times \overline{\mathrm{i}}=\overline{\mathrm{j}} \times \overline{\mathrm{j}}=\overline{\mathrm{k}} \times \overline{\mathrm{k}}=0\)

Examples Of Cross Product:

1. Torque (Or) Moment Of Force (\((\bar{\tau})\)): It is defined as the product of force and perpendicular distance from the point of application.

∴ Torque \(\tau=\overline{\mathrm{r}} \times \overline{\mathrm{F}}\)

2. Angular Momentum And Angular Velocity: For a rigid body in motion, Angular momentum \((\overline{\mathrm{L}})\) = radius \((\overline{\mathrm{r}})\) x momen- turn \((\overline{\mathrm{p}})\)

∴ Angular momentum \((\overline{\mathrm{L}})\)

= \(\overline{\mathrm{r}} \times(\mathrm{m} \overline{\mathrm{v}})=\mathrm{m}(\overline{\mathrm{r}} \times \overline{\mathrm{v}})\)

Question 5. Define angular velocity. Derive v = rω.
Answer:

Angular Velocity (ω): The rate of change of angular displacement is called angular velocity.

Relation Between Linear Velocity (V) And Angular Velocity (ω): Let a particle P is moving along the circumference of a circle of radius V with uniform speed v.

Let it be initially at position A, during a small time Δt it goes to a new position say C from B. The Angle subtended during this small interval is say dθ.

By definition angular velocity, \(\omega=\mathrm{Lt}_{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta \mathrm{t}}=\frac{\mathrm{d} \theta}{\mathrm{dt}}\)

But linear velocity \(v=\underset{\Delta t \rightarrow 0}{\mathrm{Lt}} \frac{\mathrm{BC}}{\delta \mathrm{t}}\)

When \(\theta\) is small \(\text{arc} B C=r dθ \theta\)

⇒ v = \(\mathrm{r} \cdot \frac{\mathrm{d} \theta}{\mathrm{dt}}\)

∴ \(\mathrm{v}=\mathrm{r} \cdot \omega\) (because \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=\omega\))

Question 6. State the principle of conservation of angular momentum. Give two examples.
Answer:

Law Of Conservation Of Angular Momentum:

When no external torque is acting on a body then the angular momentum of that rotating body is constant.

i.e., \(\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2(\text { when } \tau=0)\)

Example 1: A boy stands over the centre of a horizontal platform which is rotating freely with a speed ω₁ (n₁ revolutions/sec.)  about a vertical axis passing through the centre of the platform and straight up through the boy.

He holds two bricks in each of his hands close to his body. The combined moment of inertia of the system is said l₁. Let the boy stretch his arms to hold the masses far away from his body. In this position, the moment of inertia increases to I₂ and lets ω₂ be his angular speed.

Here ω₂ <ω₁ because the moment of inertia increases.

∴ \(\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \Rightarrow \mathrm{I}_1 \frac{2 \pi}{\mathrm{T}_1}=\mathrm{I}_2 \frac{2 \pi}{\mathrm{T}_2} \Rightarrow \mathrm{I}_1 \mathrm{n}_1=\mathrm{I}_2 \mathrm{n}_2\)

Example 2: An athlete diving off a high springboard can bring his legs and hands close to the body and perform a Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall his angular momentum remains constant.

Systems of Particles and Rotational Motion KSEEB Physics Chapter 7 

Question 7. Define angular acceleration and torque. Establish the relation between angular acceleration and torque.
Answer:

Angular acceleration (α): The rate of change of angular velocity is called angular acceleration.

Torque: It is defined as the product of the force and the perpendicular distance of the point of application of the force from that point.

Relation Between Angular Acceleration And Torque:

We know that, L = Iω

On differentiating the above expression with respect to time ‘t’ \(\frac{\mathrm{dL}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{I} \omega) \Rightarrow \frac{\mathrm{dL}}{\mathrm{dt}}=I \frac{\mathrm{d} \omega}{\mathrm{dt}}\)

But \(\frac{dL}{dt}\) is the rate of change of angular momentum called “Torque(τ)”.

and \(\frac{d \omega}{d t}\) is the rate of change of angular velocity called “angular acceleration (α)”

∴ The relation between Torque and angular acceleration is τ = Iα,

Question 8. Write the equations of motion for a particle rotating about a fixed axis.
Answer:

Equations Of Rotational Kinematics: If ‘θ’ is the angular displacement, ω₁ is the initial angular velocity, ω_f is the final angular velocity after a time ‘t’ seconds and ‘α’ is the angular acceleration, then the equations of rotational kinematics can be written as,

⇒ \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\)

⇒ \(\theta=\omega_{\mathrm{i}} t+\frac{1}{2} \alpha \mathrm{t}^2\)

∴ \(\omega_i^2-\omega_{\mathrm{i}}^2=2 \alpha \theta\)

Question 9. Derive expressions for the final velocity and total energy of a body rolling without slipping.
Answer:

A rolling body has both translational kinetic energy and rotational kinetic energy. So the total K.E energy of a rolling body is,

⇒ \(\mathrm{K} \cdot \mathrm{E}_{\mathrm{T}}=\mathrm{K} \cdot \mathrm{E}_{\text {(Translational) }}+\mathrm{K} \cdot \mathrm{E}_{\text {(Rotational) }}\)

⇒ \(K \cdot E_T=\frac{1}{2} m v^2+\frac{1}{2} l_{\omega^2} \text {. }\)

⇒ \(K. E_T=\frac{1}{2} m v^2+\frac{1}{2} m k^2 \frac{v^2}{r^2}\)

(because \(\mathrm{I}=\mathrm{mk}^2\) and \(\mathrm{v}=\mathrm{r} \omega\))

Total energy \(\mathrm{E}=\frac{1}{2} m v^2\left[1+\frac{\mathrm{k}^2}{\mathrm{r}^2}\right]\)

⇒ \(v^2=\frac{2 E}{m\left[1+\frac{k^2}{r^2}\right]}\)

∴ Velocity of the body, v = \(\sqrt{\frac{2 E}{m\left[1+\frac{k^2}{r^2}\right]}}\)

Chapter 7 Systems Of Particles And Rotational Motion Long Answer Type Question

Question 1.

1. State and prove the parallel axis theorem.
2. For a thin flat circular disk, the radius of gyration about a diameter as axis is k. If the disk is cut along a diameter AB as shown into two equal pieces, then find the radius of gyration of each piece about AB.

Answer:

1. Parallel Axis Theorem: The moment of inertia of a rigid body about an axis passing through a point is the sum of the moment of inertia about a parallel axis passing through the centre of mass (I_G) and the mass of the body multiplied by the square of the distance (MR²) between the axes i.e., I = I_G + MR²

Proof: Consider a rigid body of mass M with ‘G’ as its centre of mass. I_G is the moment of inertia about an axis passing through the centre of mass. I = The moment of inertia about an axis passing through the point ‘O’ in that plane.

Let the perpendicular distance between the axes be OG = R (say)

Consider point P in the given plane. Join OP and GP. Extend the line OG and drop a normal from ‘P’ onto it as shown.

The moment of inertia about the axis passing through the centre of mass G. \(\left(\mathrm{I}_{\mathrm{G}}\right)=\Sigma \mathrm{mGP}^2 \quad \ldots. .(1)\)….(1)

M.O.I. of (he body about an axis passing through ‘O’ (I) = ∑mOP²…..(2)

From triangle O P D \(O P^2=O D^2+D P^2\)

⇒ OD = OG + GD

∴ \(O D^2=(O G+G D)^2=O G^2+G D^2+2 O G . G D\)…..(3)

From Equations (2) and (3)

I = \(\Sigma \mathrm{mOP} \mathrm{P}^2=\Sigma \mathrm{m}\left[\left(\mathrm{OG}^2+\mathrm{GD}^2\right.\right.\)

+ \(20 G . G D)+D^2 \text { ] }\)

∴ I\(=\Sigma \mathrm{m}\left\{\mathrm{GD}^2+\mathrm{DP}^2+\mathrm{OG}^2+20 \mathrm{G} . \mathrm{GN}\right\}\)…..(4)

But \(\mathrm{GD}^2+\mathrm{DP}^2=\mathrm{GP}^2\)

∴ \(\mathrm{I}=\Sigma \mathrm{m}\left\{\mathrm{GP}^2+\mathrm{OG}^2+20 \mathrm{OG} . \mathrm{GD}\right\}\)

∴ \(\mathrm{I}=\Sigma \mathrm{m} \mathrm{GP}^2+\Sigma \mathrm{mOG}^2+20 \mathrm{G} \mathrm{mGD} \text {. }\)

But the terms \(\Sigma m \mathrm{mP}^2=\mathrm{I}_{\mathrm{G}}\)

∴ \(\Sigma \mathrm{mOG}^2=\mathrm{MR}^2\) (because \(\Sigma \mathrm{m}=\mathrm{M}\) and \(O G=\mathrm{R}\))

The term 2OG ∑mGD = 0. Because it represents the sum of the moment of masses about the centre of mass. Hence its value is zero.

∴ I = I_G+ MR²

Hence parallel axis theorem is proved,

2. Moment of inertia of a disc of mass ‘M’ and radius ‘R’ about its diameter is, \(\mathrm{I}=\frac{\mathrm{MR}^2}{4}\)

If ‘k’ is the radius of the gyration of the disc then, I = Mk²∴

∴ \(\mathrm{Mk}^2=\frac{\mathrm{MR}^2}{4} \Rightarrow \mathrm{k}=\mathrm{R} / 2\)

After cutting along the diameter, the mass of each piece \(=\frac{\mathrm{M}}{2}\)

Moment of inertia of each piece, \(I^{\prime}=\frac{1}{4} \times \frac{M}{2} \times R^2\)

If \(\mathrm{k}^{\prime}\) is the radius of gyration of each piece then, \(\mathrm{I}^{\prime}=\frac{\mathrm{M}}{2}\left(\mathrm{k}^{\prime}\right)^2\)

∴ \(\frac{M}{2}\left(k^{\prime}\right)^2=\frac{1}{4} \times \frac{M}{2} \times R^2 \Rightarrow k^{\prime}=\frac{R}{2}=k\)

KSEEB Physics Class 11 Systems of Particles and Rotational Motion Notes

Question 2.

1. State and prove the perpendicular axis theorem.
2. If a thin circular ring and a thin flat circular disk of the same mass have the same moment of inertia about their respective diameters as the axis. Then find the ratio of their radii.

Answer:

1. Perpendicular Axis Theorem: The I moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moment of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.

∴ \(\mathrm{I}_z=\mathrm{I}_{\mathrm{x}}+\mathrm{I}_{\mathrm{y}}\)

Proof: Consider a rectangular plane lamina. X and Y are two mutually perpendicular axes in the plane. Choose another perpendicular axis Z passing through the point ‘O’.

Consider a particle P in the XOY plane.

Its coordinates are (x, y).

The moment of inertia of the particle about the X-axis is I_X = ∑my².

M.O.I about Y-axis is I_Y = ∑mx²

M.O.I about the Z axis is I_Z = ∑ m. OP²

From triangle OAP, \(\mathrm{OP}^2=\mathrm{OA}^2+\mathrm{AP}^2=\mathrm{y}^2+\mathrm{x}^2\)

∴ \(\mathrm{I}_{\mathrm{Z}}=\Sigma \mathrm{mOP} \mathrm{P}^2=\Sigma \mathrm{m}\left(\mathrm{y}^2+\mathrm{x}^2\right)\)

∴ \(\mathrm{I}_{\mathrm{Z}}=\Sigma \mathrm{my}^2+\Sigma \mathrm{mx}^2\)

But \(\Sigma m y^2=I_x\) and \(\Sigma m x^2=I_y\)

∴ Moment of Inertia about a perpendicular axis passing through ‘O’ is \(\mathrm{I}_z=\mathrm{I}_{\mathrm{x}}+\mathrm{I}_{\mathrm{y}}\)

Hence perpendicular axis theorem is proved.

2. Moment of inertia of a thin circular ring about its diameter is, \(I_1=m_1 R_1^2\)

The moment of inertia of a flat circular disc about its diameter is; \(\mathrm{I}_2=\frac{\mathrm{m}_2 \mathrm{R}_2^2}{2}\)

Given that two Objects have the same moment of inertia i.e., I₁ = I₂

⇒ \(m_1 R_1^2=\frac{m_2 R_2^2}{2}\)

Given that \(m_1=m_2\) hence \(R_1^2=\frac{R_2^2}{2}\)

⇒ \(\frac{R_1^2}{R_2^2}=\frac{1}{2}\) or \(\frac{R_1}{R_2}=\frac{1}{\sqrt{2}}\)

∴ Ratio of radii, \(R_1: R_2=1: \sqrt{2}\)

Question 3. State and prove the principle of conservation of angular momentum. Explain the principle of conservation of angular momentum with examples.
Answer:

Law Of Conservation Of Angular Momentum: When no external torque is acting on a body then the angular momentum of that rotating body is constant.

i.e., \(\left.\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \text { (when } \tau=0\right)\)

Explanation: Here I₁ and I₂ are the moment of inertia of rotating bodies and ω₁ and ω₂ are their initial and final angular velocities. If external torque \(\tau=0\) then \(\frac{\mathrm{d} \overline{\mathrm{L}}}{\mathrm{dt}}=\tau=0\)

∴ Change in angular momentum is also zero

⇒ \(\mathrm{d} \overrightarrow{\mathrm{L}}_2-\mathrm{d} \overrightarrow{\mathrm{L}}_1=0\)

i.e., \(\mathrm{I}_2 \omega_2-\mathrm{I}_1 \omega_1=0\) or \(\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2\)

Example 1: A boy stands over the centre of a horizontal platform which is rotating freely with a speed ω₁ (n₁ revolutions/sec.) about a vertical axis passing through the centre of the platform and straight up through the boy.

He holds two bricks in each of his hands close to his body. The combined moment of inertia of the system is say I₁. Let the boy stretch his arms to hold the masses far away from his body. In this position, the moment of inertia increases to I₂ and lets ω₂ be his angular speed.

Here ω₂ < ω₁ because the moment of inertia increases.

∴ \(I_1 \omega_1=I_2 \omega_2 \Rightarrow I_1 \frac{2 \pi}{T_1}=I_2 \frac{2 \pi}{T_2} \Rightarrow I_1 n_1=I_2 n_2\)

KSEEB Class 11 Physics Systems of Particles and Rotational Motion 

Example 2: An athlete diving off a high springboard can bring his legs and hands close to the body and perform a Somersault about a horizontal axis passing through his body in the air before reaching the water | below it. During the fall his angular momentum remains constant.

Position Of Centre Of Mass Of Some Symmetrical Bodies:

Comparison Of Translatory And Rotatory Motions:

KSEEB Class 11 Physics Systems Of Particles And Rotational Motion Problems

Question 1. Show that a (b x c) is equal in magnitude to the volume of the parallelopiped formed on the three vectors a, b and c.
Solution:

Let a parallelopiped be formed on three vectors \(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c}\)

Now, \(\vec{b} \times \vec{c}\)= bc \(\sin 90^{\circ} \hat{n}\) = bc \(\hat{n}\)

where \(\hat{n}\) is the unit vector along \(\overrightarrow{O A}\) perpendicular to the plane containing \(\vec{b}\) and \(\vec{c}\)

Now \(\vec{a} \cdot(\vec{b} \times \vec{c})\) = \(\vec{a} \cdot b c \hat{n}\) =(a)(b c) \(\cos 0^°=a b c\)

Which is equal in magnitude to the volume of parallelopiped.

Question 2. A rope of negligible mass is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope? Assume that there is no slipping.
Solution:

Here M = 3 kg ; R = 40 cm = 0.4 m

Moment of inertia of the hollow cylinder about its axis, I = MR² = 3(0.4)² = 0.48 kg m²

Force applied, F = 30 N

∴ Torque, τ = F x R =30×0.4 =12N-m

If α is angular acceleration produced, then from τ = Iα

⇒ \(\alpha=\frac{\tau}{1}=\frac{12}{0.48}=25 \mathrm{rad} \mathrm{s}^{-2}\)

Linear acceleration, a = Rα = 0.4 x 25 = 10 ms^-2.

Question 3. A coin is kept at a distance of 10 cm from the centre of a circular turn table. If the coefficient of static friction between the table and the coin is 0.8. Find the frequency or rotation of the disc at which the coin will just begin to slip.
Solution:

Distance of coin = r = 10 cm = 0.1 m.

Coefficient of friction μ = 0.8.

Frequency of rotation = number of rotations per second.

For coin not to slip \(\mu \mathrm{mg}=\mathrm{mr} \omega^2 \Rightarrow \omega=\sqrt{\frac{\mu g}{\mathrm{r}}}\)

∴ \(\omega=\sqrt{\frac{0.8 \times 9.8}{1}}=\sqrt{78.4}=8.854 \mathrm{rad} / \mathrm{sec} \text {. }\)

Number of rotations per \(\mathrm{sec}=\)

Rotational frequency \(=\frac{\omega}{2 \pi}=\frac{8.854}{2 \pi}=1.409 \text{Rot} / \mathrm{sec}\) or 84.54 r.p.m.

Question 4. Find the torque of a force \(7 \overrightarrow{\mathrm{i}}+3 \overrightarrow{\mathrm{j}}-5 \overrightarrow{\mathrm{k}}\) about the origin. The force acts on a particle whose position vector is \(\overrightarrow{\mathbf{i}}-\overrightarrow{\mathbf{j}}+\overrightarrow{\mathbf{k}}.\).
Solution:

Force \(\overrightarrow{\mathrm{F}}=7 \overrightarrow{\mathrm{i}}+3 \overrightarrow{\mathrm{j}}-5 \overrightarrow{\mathrm{k}}\);

Position of particle \(\vec{r}=\vec{i}-\vec{j}+\vec{k}\)

Torque \(\mathrm{t}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}\)

∴ \(\tau=\left|\begin{array}{rrr}
\overrightarrow{\mathrm{i}} & \vec{j} & \vec{k} \\
1 & -1 & 1 \\
7 & 3 & -5
\end{array}\right|\)

∴ \(\tau= \overrightarrow{\mathrm{i}}[(-1)(-5)-(3 \times 1)]\)–\(\overrightarrow{\mathrm{j}}[1 \times(-5)-(7 \times 1)]\)

+ \(\overrightarrow{\mathrm{k}}[(1 \times 3)-(-1) \times 7]\)

⇒ \(\tau=\vec{i}(5-3)-\vec{j}(-5-7)+\vec{k}(3+7)\)

= \(2 \overrightarrow{\mathrm{i}}+12 \overrightarrow{\mathrm{j}}+10 \overrightarrow{\mathrm{k}}\)

Magnitude of torque \(\tau=\sqrt{2^2+12^2+10^2}\)

= \(\sqrt{4+144+100}=\sqrt{248}\) units

Question 5. Particles of masses 1g, 2g, 3g…, and 100g are kept at the marks 1 cm, 2 cm, 3 cm…, and 100 cm respectively on a meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale.
Solution:

Given Masses of lg, 2g, 3g 100 g are 1, 2, 3…….100 cm on a scale.

1. Sum of masses \(\Sigma m=\sum_{i=1}^n n i\)

The sum of n natural numbers S = \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{100 \times 101}{.2}=5051\)

∴ Total mass \(\mathrm{M}=5051 \mathrm{gr}=5.051 \mathrm{~kg} \rightarrow\) (1)

2. The centre of mass of all these masses is given by

⇒ \(X_{C. M}=\frac{\left(m_1 x_1+m_2 x_2+\ldots . .\right)}{\Sigma m}\)

= \(\frac{1 \times 1+2 \times 2+3 \times 3+\ldots .+100 \times 100}{M}\)

= \(\frac{1^2+2^2+3^2+\ldots .+100^2}{M}\)

But sum of squares of 1st n natural numbers is S = \(\frac{n(n+1)(2 n+1)}{6}\)

∴ CM = \(\frac{n(n+1)(2 n+1)}{6 \times n(n+1)}\) use \(M=\frac{n(n+1)}{2}\)

∴ CM = \(\frac{201}{3}=67 C M \quad \rightarrow \text { (2) }\)

3. M . O .1 .=1

= \(m_1 r_1^2+m_2 r_2^2+m_3 r_3^2+\ldots . .+m_{100} r_{100}^2\)

∴ I = \(1 \times 1 \times 1+2 \times 2^2+3 \times 3^2+\ldots \ldots \ldots\) + \(100 \times 100^2\)

= \(1+2^3+3^3+\ldots \ldots+100^3 \times 10^{-7} \mathrm{Kg} \cdot \mathrm{m}^2\)

(because 1g = \(10^{-3} \mathrm{~kg} \ 1 . C M=10^{-2} \mathrm{~m}\))

Sum of cubes of 1st n natural numbers is S = \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\)

I = \(\frac{100^2 \times(101)^2}{4} \mathrm{~g} \cdot \mathrm{cm}^2\)

= \(2.550 \times 10^7 \mathrm{~g} \cdot \mathrm{cm}^2=2.550 \mathrm{Kg} \cdot \mathrm{m}^2\) (From one end of scale).

M.O.I. about C.M. = \(I_G=I-\mathrm{MR}^2\)

= \(2.550-5.05 \times 0.67 \times 0.67\)

= \(2.550-2.267=0.283 \mathrm{~kg} \cdot \mathrm{m}^2\)

4. Perpendicular bisector is at 50 CM.

So shift M.O.I from centre of mass to x₁ = 50cm point from x = 67 CM

∴ Distance between the axis R = 67 – 50 = 17cm = 0.17M

M.O.I. about this axis I = I_G + MR²

= 0.283 + 5.05 x 0.17 x 0.17 = 0.283 + 0.146 = 0.429 kgm²

M.O.I. about perpendicular bisector of scale = 0.429 kg – m_²

Systems of Particles And Rotational Motion Questions And Answers KSEEB Physics

Question 6. Calculate the moment of inertia of a flywheel, if its angular velocity is increased from 60 r.p.m. to 180 r.p.m. when 100 J of work is done on it.
Solution:

W=\(100 \mathrm{~J}, \omega_1=60 \text { RPM }=1 \text { R.P.S }=2 \pi \text { Rad. } \\\), \(\omega_2=180 \text { R.P.M. }=3 \text { R.P.S }=6 \pi \text { Rad. }\)

Work done, \(\mathrm{W}=100 \mathrm{~J}=\frac{1}{2} \mathrm{I} \omega_2^2-\frac{1}{2} \mathrm{I} \omega_1^2\)

∴ 100 = \(\frac{1}{2} I\left(36 \pi^2\right. \left.-4 \pi^2\right) \Rightarrow I=\frac{200}{32 \pi^2}\)

= \(0.6332 \mathrm{kgm} \mathrm{m}^2=0.63 \mathrm{~kg} \cdot \mathrm{m}^2\)

Question 7. Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid of the triangle and perpendicular to its plane.
Solution:

Mass of each particle m = 100 g; side of equilateral triangle = 10 cm.

In equilateral triangle height of angular bisector \(\mathrm{CD}=\frac{\sqrt{3}}{2} l\)

Centroid will divide the angular bisector in a ratio 2 :1 So X distance of each mass from vertex to centroid is \(\text { 2. } \frac{\sqrt{3}}{3} l=\frac{1}{\sqrt{3}} l\)

Moment of Inertia of the system

I = \(m_1 r_1^2+m_2 r_2^2+m_3 r_3^2\)

∴ I = \(0.1 \times\left(\frac{\sqrt{3}}{3} \times 0.1\right)^2+0.1 \times\left(\frac{\sqrt{3}}{3} \times 0.1\right)^2\)

+ \(0.1 \times\left(\frac{\sqrt{3}}{3} \times 0.1\right)^2\)

= \(3 \times \frac{3}{9} \times 0.1^3=1 \times 10^{-3} \mathrm{kg.} \mathrm{m}\)

Question 8. Four particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.
Solution:

Mass of each particle, m = 100 g = 0.1 kg.

Length of side of square, a = 10 cm = 0.1 m

In square distance of corner from centre of square = – diagonal = \(\frac{\sqrt{2} a}{2}=\frac{a}{\sqrt{2}}\)

∴ Total moment of Inertia

1. I = \(m_1 r_1^2+m_2 r_2^2+m_3 r_3^2+m_4 r_4^2=4 \mathrm{mr}^2\)

I = \(4 \times 0.1\left(\frac{1}{\sqrt{2}} \times 0.1\right)^2=4 \times \frac{1}{2} \times 0.1^3\)

= \(2 \times 10^{-3} \mathrm{Kg} \cdot \mathrm{m}^2\)

2. Radius of gyration \(\mathrm{k}=\sqrt{\frac{1}{\mathrm{~m}}}=\sqrt{\frac{2 \times 10^{-3}}{4 \times 10^{-1}}}\)

= \(\frac{1}{\sqrt{2}} \times 10^{-1}\)

= \(0.7071 \times 10^{-1} \mathrm{~m} \text { or } 7.071 \mathrm{~cm} .\)

Question 9. Two uniform circular discs, each of mass 1 kg and radius 20 cm, are kept in contact about the tangent passing through the point of contact. Find the moment of inertia of the system about the tangent passing through the point of contact.
Solution:

Mass of disc = M = 1 kg.

Radius of disc = 20 cm = 0.2 m

They are in contact as shown.

M.O.I of a circular disc about a tangent parallel to its plane = \(\frac{5}{9}\) MR²

Total M.O.I of the system

I = \(\frac{5}{4} M R^2+\frac{5}{4} M R^2=\frac{5}{2} M R^2\)

∴ \(\mathrm{I}=\frac{5}{2} \times 1 \times 0.2 \times 0.2=5 \times 1 \times 0.1 \times 0.2\)

= \(0.1 \mathrm{kgm}^2\)

Question 10. Four spheres each diameter 2a and mass ‘m’ are placed with their centres on the four corners of a square of the side b. Calculate the moment of inertia of the system about any side of the square.
Solution:

Diameter of sphere = 2a ⇒ radius = a.

Side of square = b.

For spheres 1 and 2 axis of rotation is the same and passes through diameters.

∴ M.O.I. of spheres \(1 \ 2=\mathrm{I}_1+\mathrm{I}_2=\frac{2}{5} \mathrm{ma}^2\)

+ \(\frac{2}{5} m a^2=\frac{2}{5} m a^2 \rightarrow 1\)

For spheres 3,4 M.O.L about diameters

= \(\frac{2}{5} \mathrm{ma}^2+\frac{2}{5} \mathrm{ma}^2=\frac{2}{5} \mathrm{ma}^2\)

Transfer this M.O.I. onto the axis using the Parallel axis theorem.

∴ \(\mathrm{I}_3+\mathrm{I}_4=\frac{2}{5} m a^2+m b^2+\frac{2}{5} m a^2+m b^2\)

= \(\frac{4}{5} m a^2+2 m b^2\)

Total M.O.I. of the system

= \(\frac{4}{5} m a^2+\frac{4}{5} m a^2+2 m b^2\)

∴ I = \(\left[\frac{8}{5} m a^2+2 m b^2\right]\)

Question 11. To maintain a rotor at a uniform angular speed or 200 rad s^-1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution:

Given uniform angular speed (ω) = 200 rad s^-1

Torque, τ = 180 N-m;

But power p = τω =

∴ P = 180 x 200 = 36000 watt = 36 kW

Question 12. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Solution:

Let m be the mass of the stick concentrated at C, the 50 cm mark

For equilibrium about C’, i.e. at the 45 cm mark, 10g (45-12) = mg (50-45)10g x 33 = mg x 5

m = \(\frac{10 \times 33}{5}\) = 66 grams

Question 13. Determine the kinetic energy of a circular disc rotating with a speed of 60 rpm about an axis passing through a point on its circumference and perpendicular to its plane.
The circular disc has a mass of 5 kg and a radius of 1 m. 
Solution:

Mass of disc, M = 5 kg; Radius R = 1 m.

Angular velocity, \(\omega=60 \mathrm{RPM}=\frac{60 \times 2 \pi}{60}\)

= \(2 \pi \mathrm{Rad} / \mathrm{sec}\).

M.O.I. of the disc about a point passing through the circumference and perpendicular to the plane. \(I^{\prime}=I_G+M R^2\)

But \(\mathrm{I}_{\mathrm{G}}=\frac{\mathrm{MR}^2}{2}\) (for above case)

∴ \(\mathrm{I}^{\prime}=\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2\)

Rotational Kinetic Energy = \(\frac{1}{2} I \omega^2\)

= \(\frac{1}{2} \cdot \frac{3}{2} \mathrm{MR}^2 \times(2 \pi)^2\)

∴ R.K.E. = \(\frac{3}{4} \times 5 \times 1 \times 1 \times 2 \times 3.142 \times 2 \times 3.142\)

= \(15 \times 3.142 \times 3.142=148.1 \mathrm{~J} \text {. }\)

Question 14. Two particles, each of mass m and speed u, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two-particle system is the same whatever the point about which the angular momentum is taken.
Solution:

Angular momentum, L = mvr

Choose any axis say ‘A’

Let at any given time distance between m₁ and m₂ = L = L₁ + L₂

About the axis ‘A’ both will rotate in the same direction.

∴ Total angular momentum \(\mathrm{L}=\mathrm{L}_1+\mathrm{L}_2=m u L_1+m u L_2=m u\left(L_1+L_2\right)\) = muL

about any new axis say B distance of m₁ and m₂, are say \(\mathrm{L}_1^{\prime} \text { and } \mathrm{L}_2^{\prime}\)

Total angular momentum, \(L=m u L_1^{\prime}+m u L_2^{\prime}\)

or \(L=m u\left(L_1^{\prime}+L_2^{\prime}\right)=m u L\)

(because \(L_1^{\prime}+L_2^{\prime}=L\))

Hence, the total angular momentum of the system is always constant.

Question 15. The moment of inertia of a flywheel making 300 revolutions per minute is 0.3 kgm². Find the torque required to bring it to rest in the 20s.
Solution:

M.O.I, I = 0.3 kg. ; time, t = 20 sec.,

ω₁ = 300 R.P.M. = \(\frac{300}{60}\) = 5. R.P.S.; ω₂ = 0

Torque, \(\tau=\mathrm{l} \alpha=0.3 \times \frac{5 \times 2 \pi}{20}=0.471 \mathrm{~N}-\mathrm{m}\)

Question 16. When 100J of work is done on a flywheel, its angular velocity is increased from 60 rpm to 180 rpm. What is the moment of inertia of the wheel?
Solution:

W = 100J, ω₁ = 60 RPM = 1 R.P.S = 2π Rad.

ω₂ = 180 R.P.M. = 3 R.P.S = 6π Rad.

Work done, \(W=100 \mathrm{~J}=\frac{1}{2} \mathrm{I} \omega_2^2-\frac{1}{2} \mathrm{I} \omega_1^2\)

∴ 100 = \(\frac{1}{2} \mathrm{I}\left(36 \pi^2-4 \pi^2\right) \Rightarrow I=\frac{200}{32 \pi^2}\)

= \(0.6332 \mathrm{kgm} \mathrm{m}^2=0.63 \mathrm{~kg} \cdot \mathrm{m}^2\)

Question 17. Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g and 200g respectively. Each side of the equilateral triangle is 0.5 m long, 100g mass is at the origin and 150g mass is on the X-axis.
Solution:

Mass at A = 100g ; Coordinates = 0, 0

Mass at B = 150 g; Coordinates = (0.5, 0)

Mass at C = 200g; Coordinates (0.25,0.25 √3)

Coordinates \(x_{c m}=\frac{m_A x_A+m_B x_B+m_C x_C}{m_A+m_B+m_C}\)

= \(\frac{(100 \times 0)+(150 \times 0.5)+(200 \times 0.25)}{100+150+200}\)

= \(\frac{75+50}{450}=\frac{125}{450}=\frac{5}{18} \mathrm{~m}\)

∴ \(Y_{c m}=\frac{m_A y_A+m_B y_B+m_C y_C}{m_A+m_B+m_C}\)

= \(\frac{(100 \times 0)+(150 \times 0)+(200 \times 0.25 \sqrt{3})}{100+150+200}\)

= \(\frac{50 \sqrt{3}}{450}=\frac{\sqrt{3}}{9}=\frac{1}{3 \sqrt{3}} \mathrm{~m}\)

Question 18. Find the scalar and vector product of two vectors \(\vec{a}=(3 \hat{i}-4 \hat{j}+5 \hat{k})\) and \(\overrightarrow{\mathbf{b}}=(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \text {. }\)
Solution:

⇒ \(a \cdot b=(3 \hat{i}-4 \hat{j}+5 \hat{k}) \cdot(-2 \hat{i}+\hat{j}-3 \hat{k})\)

= -6-4-15=-25

⇒ \(\mathbf{a} \times \mathbf{b}\) = \(\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & -4 & 5 \\
-2 & 1 & -3
\end{array}\right|\)

= \(7 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}}\)

∴ \(\mathrm{b} \times \mathrm{a}=-7 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}} \) (Recognise.)

KSEEB Class 11 Physics Solutions For Chapter 6 Work, Energy, and Power Important Points

Dot Product (Or) Scalar Product: The scalar product (or) dot product of any two vectors A and B is \(\bar{A} \text { and } \bar{B} \text { is } \bar{A} \cdot \bar{B}=|\bar{A}||\bar{B}| \cos \theta\)

where θ is the angle between them.

Dot Product (Or) Scalar Product Example: Work W = \(\bar{F} \cdot \bar{S}\)

Read and Learn More KSEEB Class 11 Physics Solutions

Properties Of Dot Product:

1. Scalar product obeys “commutative law” i. e., \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}=\overline{\mathrm{B}} \cdot \overline{\mathrm{A}}\)
2. Scalar product obeys “distributive law” \(\overline{\mathrm{A}} \cdot(\overline{\mathrm{B}}+\overline{\mathrm{C}})=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}+\overline{\mathrm{A}} \cdot \overline{\mathrm{C}}\)
3. In unit vector i, j, k system \(\overline{\mathrm{i}} \cdot \overline{\mathrm{i}}=\overline{\mathrm{j}} \cdot \overline{\mathrm{j}}=\overline{\mathrm{k}} \cdot \overline{\mathrm{k}}=1\) i.e., the dot product of like unit vectors is unity.
   1. \(\overline{\mathrm{i}} \cdot \overline{\mathrm{j}}=\overline{\mathrm{j}} \cdot \overline{\mathrm{k}}=\overline{\mathrm{k}} \cdot \overline{\mathrm{i}}=0\) i.e., dot product of perpendicular vectors is zero.

Example: If \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}=0 \Rightarrow \overline{\mathrm{A}} \text { and } \overline{\mathrm{B}}\) are pendicular.

Work: Work done by a force is defined as the product of a component of force in the direction of displacement and the magnitude of displacement.

W = \(\overline{\mathrm{F}} \cdot \overline{\mathrm{S}} \text { (or) } \mathrm{W}=\overline{\mathrm{F}} \cdot \overline{\mathrm{S}} \cos \theta\)

Work is a scalar, unit kg² /sec² called joule (J), D.F = ML²T^-2

Energy: The ability (or) capacity of a body to do work is called energy.

Work, Energy, And Power Questions And Answers KSEEB Physics

Note: Energy can be termed as stored work in the body.

Kinetic Energy: Energy possessed by a moving body is called kinetic energy (k).

KE = \(\frac{1}{2}\) mv²

Kinetic Energy Example: All moving bodies contain kinetic energy.

Relation Between Kinetic Energy And Momentum: Kinetic energy KE = \(\frac{1}{2}\) mv  momentum \(\overline{\mathrm{p}}=\mathrm{mv}\)

∴ E = \(\frac{P^2}{2 m}\)

Work Energy Theorem (For Variable Force): Work done by a variable force is always equal to the change in kinetic energy of the body.

Work done \(\mathrm{W}=\frac{1}{2} \mathrm{mV}^2-\frac{1}{2} \mathrm{mV}_0^2=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}\)

Potential Energy (V): It is the energy possessed by virtue of the position (or) configuration of a body.

Law Of Conservation Of Mechanical Energy: The total mechanical energy of a system is conserved if the forces doing work on it are conservative.

Spring Constant (k): It is defined as the ratio of force applied to the displacement produced in the spring.

k = \(-\frac{\mathrm{F}_{\mathrm{s}}}{\mathrm{x}}\) unit: Newton/metre, D.F = MT^-2

Note: A spring is said to be stiff if k is large. A spring is said to be soft if k is small.

Law Of Conservation Of Energy: When forces doing work on a system are conservative then the total energy of the system is constant i. e. energy can neither be created nor destroyed.

Law Of Conservation Of Energy Explanation: When conservative forces are applied to a system then the total mechanical energy is

1. As kinetic energy which depends on motion OR
2. As potential energy which depends on the position of the body.

Collisions: In collisions, a moving body collides with another body. During collision, the two colliding bodies are in contact for a very small period. During time of contact the two bodies will exchange the momentum and kinetic energy.

Collisions Are Two Types:

1. Elastic collision
2. Inelastic collision.

1. Elastic collisions: Elastic collisions will obey
   • Law of conservation of momentum and
   • Law of conservation of energy.
2. Inelastic Collisions: Inelastic collisions will follow, the law of conservation of momentum only.

In these collisions, a part of energy is lost in the form of heat, sound etc.

Coefficient Of Restitution (e): The coefficient of restitution is the ratio or relative velocity of separation (v₂– v₁) to the relative velocity of approach (u₁-u₂).

Coefficient of restitution e = \(\frac{v_2-v_1}{u_1-u_2}\)

1. For Perfect elastic collisions e = 1
2. For Perfect inelastic collisions e = 0
3. For collisions ’e’ lies between ’0’ and ‘1’.

One-Dimensional Collision (Or) Head-On Collisions: If the two colliding bodies are moving along the same straight line they are called dimensional collisions or hear-on collisions.

For these collisions, the initial and final velocities of the two colliding bodies are along the same straight line.

Two-Dimensional Collisions: If the two bodies moving in a plane collide and even after collision they are moving in the same plane, then such collisions are called two-dimensional collisions.

Power (P): It is the rate of doing work.

Power (P) = \(\frac{\text { work }}{\text { time }}\), Unit: Watt

Dimensional formula \(\mathrm{ML}^2 \mathrm{~T}^{-3}\)

Power can also be expressed as \(\mathrm{P}=\frac{\mathrm{dw}}{\mathrm{dt}}=\overline{\mathrm{F}} \cdot \frac{\mathrm{d} \overline{\mathrm{r}}}{\mathrm{dt}}=\overline{\mathrm{F}} \cdot \overline{\mathrm{V}}\)

Kilo Watt Hour (K.W.H): If work done is at a rate of 1000joules/sec continuously for a period of one hour then that amount of work is defined as K.W.H.

K.W.H is taken as 1 unit for supplying electrical energy.

Horse Power(H.P): 746 watts is called one horsepower.

∴ 1 H.P = 746 Watt.

KSEEB Class 11 Physics Chapter 6 Work, Energy, And Power

Work, Energy, And Power Solutions KSEEB Class 11 Physics Important Formulae

Work is the product of force and displacement in the same direction.

1. When force and displacement are in same direction work, W = F.S.; Unit: joule; D.F. = ML²T^-2
2. When a force (\(\bar{F}\)) is applied with some angle ‘0’ with displacement vector (s) then work W = F • S cos θ or W = F • S
3. When a variable force is applied to a body, W = ∫F • dx

Power (P) is defined as the rate of doing work. It is a scalar.

Power, (P) = \(\frac{\text { work }}{\text { time }}=\frac{W}{t}\); Unit : watt; D.F. \(=\mathrm{ML}^2 \mathrm{~T}^{-3}\).

Work and energy can be interchanged. In machine gun problems work done = kinetic energy stored in bullets

i.e. W = \(\frac{1}{2}\) mv²

In motor and lift problems work done = change in potential energy (mgh)

Potential energy, (PE) = mgh;

Kinetic energy, KE = \(\frac{1}{2}\) mv²

Note: Work, P.E and K.E are scalars. Their units and Dimensional formulae are the same.

Relation Between KE And Momentum Are:

1. Kinetic energy, KE = \(\frac{p^2}{2m}\)
2. Momentum, \(p=\sqrt{2 m \cdot K E}\)

For a conservative force total work done in a closed path is zero. Example: Gravitational force.

For a non – conservative force work done in a closed path is not equal to zero. Example: Frictional force.

Work Energy Theorem: Work done by an unbalanced force is equal to the difference in kinetic energy.

W = \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mu²

From the Law of conservation of energy change in potential energy is equal to work done.

∴ W = mgh₂ – mgh₁

1. In elastic collisions Relative Velocity of approach = relative velocity of separation ⇒ u₁ – u₂ = v₂ — v₁
2. In the case of elastic collision, the Law of conservation of momentum and the Law of conservation of Energy are conserved. ⇒ \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)

∴ \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

In elastic collisions final velocities after collisions are

⇒ \(v_1=\left[\frac{m_1-m_2}{m_1+m_2}\right] u_{1^{+}}\left[\frac{2 m_2}{m_1+m_2}\right] u_2\)

⇒ \(v_2=\left[\frac{2 m_1}{m_1+m_2}\right] u_{1^{+}}\left[\frac{m_2-m_1}{m_1+m_2}\right] u_2\)

In perfectly Inelastic collision common velocity of the bodies, (v) = \(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\)

Coefficient of restitution, \(\mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}=\frac{\text { Velocity of separation }}{\text { Velocity of approach }}\)

For a body dropped from a height ‘h’

1. Velocity of approach, \(\mathrm{u}=\sqrt{2 \mathrm{gh}}\)
2. Coefficient of restitution, e = \(\sqrt{\frac{h_2}{h_1}}\)
3. Velocity of separation, v₁ = – e \(\mathrm{u}=\sqrt{2 \mathrm{gh}}\)
4. The height attained after the nth bounce, h_n = e²ⁿ h

KSEEB Class 11 Physics Chapter 6 Work Energy And Power Very Short Answer Questions

Question 1. If a bomb at rest explodes Into two pieces, the pieces must travel In opposite directions. Explain.
Answer:

Explosion is due to internal forces. In the law of conservation of linear, momentum internal forces cannot change the momentum of the system. So after explosion m₁v₁ + m₂v₂ = 0 or m₁v₁ = – m₂v₂. ⇒ they will fly in opposite directions.

Question 2. State the conditions under which a force does no work.
Answer:

When a force (F) and displacement (S) are mutually perpendicular then work done is zero.

∴ \(\mathrm{W}=\overline{\mathrm{F}} \cdot \overline{\mathrm{S}}=|\mathrm{F}||\mathrm{S}| \cos \theta \text { when } \theta=90^{\circ}\) work W = 0

Even though force is applied if displacement is zero then work done W = 0.

Question 3. Define Work, Power and Energy. State their S.I. units.
Answer:

Work: The product of force and displacement along the direction of force is called work.

Work done \(\mathrm{W} =\overline{\mathrm{F}} \cdot \overline{\mathrm{S}}\)

= \(|\overline{\mathrm{F}}||\overline{\mathrm{S}}| \cos \theta\)

S.I.’s unit of work is Joule.

Dimensional formula: ML²T^-2.

Power: The rate of doing work is called power.

Power = \(\frac{\text { Work done }}{\text { time }}=\frac{W}{t}\)

S.I. unit: Watt

D.F.: \(\mathrm{ML}^2 \mathrm{~T}^{-3}\)

Energy: It is the capacity or ability of the body to do work. By spending energy we can do work or by doing work energy content of the body will increase.

S.I. unit: Joule;

D.F.: ML²T^-2

KSEEB Class 11 Physics solutions for Chapter 6 Work, Energy, and Power

Question 4. Slate the relation between the kinetic energy and momentum of a body.
Answer:

Kinetic Energy, Momentum Relation:

Kinetic energy K.E = \(\frac{1}{2}\) mv²……..(1)

Momentum \(\bar{p}\) = mv…..(2)

from equation K.E = \(\frac{1}{2}\) mv²

multiply with \(\frac{m}{m}\)

K.E = \(\frac{1}{2} m v^2 \times \frac{\mathrm{m}}{\mathrm{m}}\)

= \(\frac{1}{2} \frac{\left(m^2 v^2\right)}{m}\) from equation (2) \(\bar{p}=m v\)

K.E = \(\frac{1}{2} \cdot \frac{\mathrm{p}^2}{\mathrm{~m}}\)

∴ K.E. = \(\frac{p^2}{2 m} \text { (Or) } p=\sqrt{(\text { K.E.) } 2 m}\)

Question 5. State the sign of work done by a force in the following.

1. Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
2. Work done by the gravitational force in the above case.

Answer:

1. When a bucket is lifted out of the well work is done against gravity so work done is negative.
2. Work done by the gravitational force is positive.

Question 6. State the sign of work done by a force in the following.

1. Work done by friction on a body sliding down an inclined plane.
2. Work done gravitational force in the above case.

Answer:

1. Work done by friction while sliding down is negative because it opposes the downward motion of the body.
2. Work done by gravitational force when a body is sliding down is positive.

Question 7. State the sign of work done by a force in the following.

1. Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
2. Work done by the resistive force of air on a vibrating pendulum In bringing It to rest.

Answer:

1. Work done against the direction of motion of a body moving on a horizontal plane is negative.
2. In pendulum a ∝ -y. So work done by air resistance to bring it to rest is considered as positive.

Question 8. State if each of the following statements is true or false. Give reasons for your answer.

1. The total energy of a system is always con¬served, no matter what internal and external forces on the body are present.
2. The work done by Earth’s gravitational force in keeping the moon in its orbit for its one revolution is zero.

Answer:

1. The law of conservation of energy states that energy can be neither created nor destroyed. This rule is applicable to internal forces and also to external forces when they are conservative forces,
2. Gravitational forces are conservative forces. Work done by conservative force around a closed path is zero.

Question 9. Which physical quantity remains constant

1. In an elastic collision
2. In an inelastic collision?

Answer:

1. In Elastic Collision: Momentum (P) and Energy (K.E.) are conserved, (remain constant)
2. In Inelastic Collision: only momentum is conserved, (remains constant)

Question 10. A body freely falling from a certain height h, after striking a smooth floor rebounds and h rises to a height h/2. What is the coefficient of restitution between the floor and the body?
Answer:

Given that, h₁ = h and h₂ = \(\frac{h_2}{2}\)

We know that the coefficient of restitution

e  \(=\sqrt{\frac{h_2}{h_1}}=\sqrt{\frac{\left(\frac{h}{2}\right)}{h}}=\frac{1}{\sqrt{2}}\)

∴ Coefficient of restitution, \(e=\frac{1}{\sqrt{2}}\)

Work, Energy, and Power solutions KSEEB Class 11 Physics

Question 11. What is the total displacement of a freely falling body, after successive rebounds from the same place of ground, before it comes to a stop? Assume that ‘e’ is the coefficient of restitution between the body and the ground.
Answer:

The total displacement of a freely falling body, after successive rebounds from the same place of ground, before it comes to a stop is equal to the height (h) from which the body is dropped.

KSEEB Physics Class 11 Work Energy And Power Short Answer Questions

Question 1. What is potential energy? Derive an expression for the gravitational potential energy.
Answer:

Potential Energy: It is the energy possessed by a body by virtue of its position.

Example: Energy stored in water an overhead tank, wound spring

Equation For Potential Energy: Let a body of mass m be lifted through a height h’ above the ground. Where the ground is taken as a reference. In this process, we are doing some work.

Work done against gravity W = m.g.h.

i.e. Force x displacement along the direction of force applied. This work is stored in the body in the form of potential energy because work and energy can be interchanged.

Potential Energy P.E. = mgh.

Question 2. A lorry and a car moving with the same momentum are brought to rest by the application of brakes, which provide equal retarding forces. Which of them will come to rest in a shorter time? Which will come to rest in less distance?
Answer:

Momentum (\(\bar{p}\) = mv) is the same for both lorry and car.

Work done to stop a body = Kinetic energy stored

∴ W = F. S = \(\frac{1}{2}\) mv² = K.E.

But the force applied by brakes is the same for lorry and car.

Relation between \(\bar{P}\) on K.E. is \(\frac{p^2}{2m}\) = F.S

or stopping distance S = \(\frac{\overline{\mathrm{p}^2}}{2 \mathrm{mF}}\)

∴ \(\overline{\mathrm{P}}\) and \(\overline{\mathrm{F}}\) are same \(\mathrm{S} \propto \frac{1}{\mathrm{~m}}\)

So lighter body (car) will travel longer distances when P and F are the same.

Then mS = constant.

∴ So a car travels a longer distance than a lorry before it is stopped.

Question 3. Distinguish between conservative and non-conservative forces with one example each.
Answer:

Conservative Forces: If work done by the force around a closed path is zero and is independent of the path, then such forces are called conservative forces.

Conservative Forces Example:

1. Work done in lifting a body In gravitational Held. When the body returns to Its original position work done on it is zero. So gravitational forces are conservative forces.
2. Let a charge ‘q’ be moved in an electric field on a closed path then change in its electric potential i.e., static forces are conservative forces.

Non-Conservative Forces: For non-con-servative forces work done by a force1 around a closed path is not equal to zero j and it is dependent on the path.

Non-Conservative Forces Example:

1. Work done to move a body against friction. While taking a body between two points say A and B. We have to do work to move the body from A to B and also work is done to move the body from B to A.
2. As a result, the work done in moving the body in a closed path is not equal to zero. So frictional forces are non-conservative forces.

KSEEB Class 11 Physics Chapter 6 Work, Power And Energy

Question 4. Show that in the case of one-dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of separation after collision.
Answer:

To show the relative velocity of approach of two colliding bodies before the collision is equal to the relative velocity of separation after the collision.

Let two bodies of masses m₁, and m₂ move with velocities u₁, and u₂ along the straight line in the same direction and collide elastically.

Let their velocities after collision be v₁ and v₂.

According to the law of conservation of linear momentum \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \)

or \(m_1\left(u_1-v_1\right)=m_2\left(v_2-u_2\right) \ldots \text { (1) }\)…..(1)

According to the law of conservation of kinetic energy \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

∴ \(m_1\left(u_1^2-v_1^2\right)=m_2\left(v_2^2-u_2^2\right) \ldots\)…..(2)

Dividing equation (2) by (1) \(\frac{u_1^2-v_1^2}{u_1-v_1}=\frac{v_2^2-u_2^2}{v_2-u_2}\)

(or) \(u_1+v_1=v_2+u_2 \Rightarrow u_1-u_2=v_2-v_1 \ldots\)……(3)

i.e., relative velocity of approach of the two bodies before collision = relative velocity of separation of the two bodies after collision. So the coefficient of restitution is equal to T.

Question 5. Show that two equal masses undergoing oblique elastic collision will move at right angles after the collision if the second body is initially at rest.
Answer:

Consider two bodies possess equal mass (m) and they undergo oblique elastic collision.

Let the first body moving with initial velocity ‘u’ collide with the second body at rest.

In elastic collision, momentum is conserved. So, conservation of momentum along the X-axis yields.

mu \(=m v_1 \cos \theta_1+m v_2 \cos \theta_2\)

(i.e.) \(u=v_1 \cos \theta_1+v_2 \cos \theta_2\)…..(1)

along Y-axis

0 = \(\mathrm{v}_1 \sin \theta_1-\mathrm{v}_2 \sin \theta_2\)…..(2)

squaring and adding equations (1) and (2) we get \(u^2=v_1^2+v_2^2+2 v_1 v_2 \cos \left(\theta_1+\theta_2\right)\)…..(3)

As the collision is elastic,

Kinetic Energy (K.E.) is also conserved.

(i.e.) \(\frac{1}{2} m u^2=\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2\)

⇒ \(u^2=v_1^2+v_2^2\)……(4)

From equations (3) and (4) \(2 v_1 v_2 \cos \left(\theta_1+\theta_2\right)=0\)

As it is given that \(v_1 \neq 0\) and \(v_2 \neq 0\)

∴ \(\cos \left(\theta_1+\theta_2\right)=0 \text { or } \theta_1+\theta_2=90^{\circ}\)

The two equal masses undergoing oblique elastic collision will move at right! angles after collision, if the second body is initially at rest.

Work, Energy, and Power KSEEB Physics Chapter 6 Solutions

Question 6. Derive an expression for the height attained by a freely falling body after the ‘n’ number of rebounds from the floor.
Answer:

Let a small ball be dropped from a height ’h’ on a horizontal smooth plate. Let it rebound to a height of ’h₁’.

The velocity with which it strikes the plate \(u_1=\sqrt{2 g h}\)

The velocity with which it leaves the plate \(v_1=\sqrt{2 g h_1}\)

The velocity of the plate before and after the collision is zero i.e., u₂ = 0, v₂ = 0

Coefficient of restitution, \(\mathrm{e}=\frac{\text { relative velocity of separation }}{\text { relative velocity of approach }}\)

= \(\frac{\sqrt{2 \mathrm{gh}_2-0}}{\sqrt{2 \mathrm{gh}_1-0}}\)

∴ \(\mathrm{e}=\sqrt{\frac{\mathrm{h}_1}{h}} \text { or } \mathrm{h}_1=\mathrm{e}^2 \mathrm{~h}\)

For 2nd rebound it goes to a height \(h_2=e^2 h_1=e^2 e^2 h=e^4 h\)

For 3rd rebound it goes to a height \(h_3=e^2 h_2=e^2 e^4 h=e^6 h\)

For nth rebound height attained \(h_n=e^{2 n} h\)

Question 7. Explain the law of conservation of energy.
Answer:

Law Conservation Of Energy: When forces doing work on a system are conservative then the total energy of the system is constant i.e., energy can neither be created nor destroyed.

i.e. Total energy = (K + u) = constant form.

Explanation: Consider a body that undergoes small displacement Δx under the action of the conservative force F. According to the work-energy theorem.

Change in K.E = work done

ΔK = F(x)Δx ………(1)

but Potential energy Δu = -F(x) Δx……(2)

from (1) and (2) = ΔK = – Δu

⇒ Δ(K + u) = 0

Hence (K + u) = constant

i.e., the sum of the kinetic energy and potential energy of the body is a constant

Since the universe may be considered as an isolated system, the total energy of the universe is constant.

KSEEB Class 11 Physics Work Energy And Power Long Answer Questions

Question 1. Develop the notions of work and kinetic energy and show that it leads to work- energy theorem. State the conditions under which a force does no work.
Answer:

Work: The product of the component of force in the direction of displacement and the magnitude of displacement is called work

W = \(\overline{\mathrm{F}} \cdot \overline{\mathrm{S}}\)

When \(\overline{\mathrm{F}}\) and \(\overline{\mathrm{S}}\) are parallel W = |\(\overline{\mathrm{F}}\)| x |\(\overline{\mathrm{S}}\)|

When \(\overline{\mathrm{F}}\) and \(\overline{\mathrm{S}}\) has some angle θ between them

W = \(\overline{\mathrm{F}}\).\(\overline{\mathrm{S}}\) cos θ

Kinetic Energy: Energy possessed by a moving body is called kinetic energy (k)

The kinetic energy of an object is a measure of the work that an object can do by virtue of its motion.

Kinetic energy can be measured with equation K = \(\frac{1}{2}\)mv²

Example: All moving bodies contain kinetic energy.

Work Energy Theorem (For Variable Force): Work done by a variable force is always equal to the change in kinetic energy of the body

Work done \(\mathrm{W}=\frac{1}{2} \mathrm{mV}^2-\frac{1}{2} m V_0^2=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}\)

Proof: Kinetic energy of a body K = \(\frac{1}{2}\)mv2

The time rate of change of kinetic energy is \(\frac{\mathrm{dk}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{2} \mathrm{mv}^2\right)=\frac{1}{2} \mathrm{~m} \cdot 2 \mathrm{v} \cdot \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{mv} \cdot \frac{\mathrm{dv}}{\mathrm{dt}} \cdot\)

But \(\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{a}\)

∴ \(\frac{\mathrm{dk}}{\mathrm{dt}}\) = m a v but ma = Force {F}

∴ \(\frac{\mathrm{dk}}{\mathrm{dt}}=\mathrm{Fv}=\mathrm{F} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}\)

∴ \(\frac{\mathrm{dk}}{\mathrm{dt}}=\mathrm{F} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}\)

Or, \(\mathrm{dk}=\mathrm{F} \cdot \mathrm{dx}\)

When force is conservative force F = F(x)

∴ On integration over initial position (x,) and final position x₂

\(\int_i^f d k=\int_{x_1}^{x_2} F(x) d x\) but \(\int_{x_1}^{x_2} F(x) d x=\) work done

By variable force W

∴ \(\mathrm{W}=\int_{\mathrm{i}}^{\mathrm{t}} \mathrm{dk} \Rightarrow W=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}\)

i.e., work done by a conservative force is equal to a change in the kinetic energy of the body.

Condition for Force not to do any work.

When force (\(\bar{F}\)) and displacement (\(\bar{S}\)) are perpendicular work done is zero. i.e., when 0 = 90 then W = \(\bar{F}\) \(\bar{S}\) = 0

KSEEB Physics Class 11 Work, Energy, and Power 

Question 2. What are collisions? Explain the possible types of collisions. Develop the theory of one-dimensional elastic collision.

(OR)

What arc collisions? Explain the possible types of collisions. Show that in the case of one-dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of separation after collision.

Answer:

A process in which the motion of a system of particles changes but keeps the total momentum conserved is called a collision.

Collisions are two types:

1. Elastic
2. In-elastic.

To show the relative velocity of the approach before the collision is equal to the relative velocity of separation after the collision.

Let two bodies of masses m₁, and m₂ moving with velocities u₁, and u₂ along the same line, in the same direction collide elastically.

Let their velocities after collision are v₁ and v₂

According to the law of conservation of linear momentum \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \)

or \(m_1\left(u_1-v_1\right)=m_2\left(v_2-u_2\right)\)……(1)

According to the law of conservation of kinetic energy \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

∴ \(m_1\left(u_1^2-v_1^2\right)=m_2\left(v_2^2-u_2^2\right)\)…….(2)

Dividing equation (2) by (1)

⇒ \(\frac{u_1^2-v_1^2}{u_1-v_1}=\frac{v_2^2-u_2^2}{v_2-u_2}\)

or \(u_1+v_1=v_2+u_2 \Rightarrow u_1-u_2=v_2-v_1 \ldots \ldots\)(3)

i.e., In clastic collisions relative velocity of approach of the two bodies before collision = relative velocity of separation of the two bodies after collision.

Velocities Of Two Bodies After Elastic Collision:

To find \(v_1\), put \(v_2=u_1-u_2+v_1\) in eqn. (1)

⇒ \(m_1 u_1+m_2 u_2=m_1 v_1+m_2\left(u_1-u_2+v_1\right)\)

⇒ \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 u_1-m_2 u_2+m_2 v_1\)

⇒ \(m_1 u_1+m_2 u_2-m_2 u_1+m_2 u_2=v_1\left(m_1+m_2\right)\)

⇒ \(u_1\left(m_1-m_2\right)+2 m_2 u_2=v_1\left(m_1+m_2\right)\)

∴ \(v_1=\frac{m_1-m_2}{m_1+m_2} u_1+\frac{2 m_2}{m_1+m_2} u_2 \ldots . . \text { (4) }\)

Similarly, \(v_2=\frac{u_2\left(m_2-m_1\right)}{m_1+m_2}+\frac{2 m_1 u_1}{m_1+m_2} . . \text { (5) }\)

Question 3. State and prove the law of conservation of energy in the case of a freely falling body.
Answer:

Law Of Conservation Of Energy: Energy can neither be created nor destroyed. But it can be converted from one form into another form so that the total energy will remain constant in a closed system.

Proof: In case of a freely falling body: Let a body of mass be dropped from a height ‘H’ at point A.

Forces due to the gravitational field are conservative forces, so total mechanical energy (E=P.E + K.E.) is constant i.e., neither destroyed nor created.

The conservation of potential energy to kinetic energy  for a ball of mass m dropped from a height H

KSEEB Class 11 Physics Work, Energy, and Power key concepts

1. At point H: Velocity of body v = 0 ⇒ K = 0

Potential energy (u) = mgH where H=height above the ground

T.E = u + K = mgH….(1)

2. At point 0: i.e., just before touching the ground

A constant force is a special case of specifically dependent force F(x) so mechanical energy is conserved.

So energy at H = Energy at 0 = mgH

Proof: At point ‘0’ height h = 0

⇒ v = \(\sqrt{2 \mathrm{gH}} \text {; }\); u = 0

⇒ \(\mathrm{K}_0=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m} 2 \mathrm{gH}=\mathrm{mgH}\)

Total energy E = mgH + 0 = mgH …….(2)

3. At Any Point h: Let height above ground = h

u = mgh, Kh = \(\frac{1}{2}\) mv²

where = \(V=\sqrt{2 g(h-x)}\)

∴ The velocity of the body when it falls through a height (h-x) is \(\sqrt{2 g(h-x)}\)

∴ Total energy = mgh = \(\frac{1}{2}\) m2g(H-h)

⇒  E = mgh + mgH – mgh = mgH…….(3)

From equations 1, 2 and 3 total energy at any point is constant.

Hence, the law of conservation of energy is proved.

Conditions To Apply The Law Of Conservation Of Energy:

1. Work done by internal forces is conservative.
2. No work is done by an external force.

When the above two conditions are satisfied then the total mechanical energy of a system will remain constant.

Problems For Work, Energy, And Power KSEEB Physics Chapter 6

Question 1. A test tube of mass 10 grnniH closed with a cork of mass 1 gram contains some ether. When the test tube is heated the cork flics out under the pressure of the ether gas. The test tube is suspended horizontally by a weightless rigid bar of length 5 cm. What is the minimum velocity with which the cork should fly out of the tube, so that the test tube describes a full vertical circle about the point O. Neglect the mass of the ether.
Solution:

Length of bar, L = 5 cm, = \(\frac{5}{100}\), g = 10m/s²

For the cork not to come out minimum, the velocity at the lowest point is, v = \(\sqrt{5 \mathrm{gL}}\).

At this condition centrifugal and centripetal forces are balanced.

∴ \(\mathrm{V}_{\min }=\sqrt{5 \times \frac{5}{100} \times 10}=\frac{5}{10} \sqrt{10} \mathrm{~m} / \mathrm{s}\) or \(0.5 \sqrt{10} \mathrm{~m} / \mathrm{s}\)

Question 2. A machine gun fires 360 bullets per minute and each bullet travels with a velocity of 600 ms^-1. If the mass of each bullet is 5 gm, find the power of the machine gun.
Solution:

Number of bullets, n = 360

Time, t = 1 minute = 60s

The velocity of the bullet, v = 600 ms^-1;

Mass  of each bullet, m = 5gm = 5 x 10^-3 kg

Power of the machine gun, \(P=\frac{n\left(\frac{1}{2} m v^2\right)}{t}\)

= \(\frac{360\left(\frac{1}{2} \times 5 \times 10^{-3} \times(600)^2\right)}{60}\)

P = \(5400 \mathrm{~W}=5.4 \mathrm{KW}\)

KSEEB Class 11 Physics Chapter 6 Work, Energy, and Power 

Question 3. Find the useful power used in pumping 3425 m³ of water per hour from a well 8 in deep to the surface, supposing 40% of the horsepower during pumping is wasted. What is the horsepower of the engine?
Solution:

Mass of water pumped, m = 3425 m³ = 3425 x 10‚ kg.

Mass of 1 m³ water = 1000 kg

Depth of well d = 8 m.,

Power wasted = 40%

∴ efficiency, η = 60%

time, t = 1 hour = 3600 sec.

Useful power = \(\frac{\text { Actual work done }}{\text { time } \times \text { efficiency }}=\frac{\mathrm{mgd}}{\mathrm{t} \cdot \eta}\)

Useful power = \(\frac{3425 \times 10 \times 9.8 \times 8 \times 100}{3600 \times 60}\)

= \(124315 \text { watt, But } 1 \mathrm{H} . \mathrm{P}=746 \mathrm{~W}\)

∴ Useful power in H.P = \(\frac{124315}{746}=166.6 \text { H.P }\)

Question 4. A pump is required to lift 600 kg of water per minute from a well 25m deep and to eject it with a speed of 50 ms^-1. Calculate the power required to perform the above task. (g = 10 m sec^-2)
Solution:

Mass of water m = 600 kg; depth = h = 25 m

Speed of water v = 25 m/s; g = 10 m/s², time t = 1 min = 60 sec.

Power of motor P = Power to lift water (P₁) + Kinetic energy of water (K.E) per second.

Power to lift water \(P_1=\frac{\mathrm{mgh}}{\mathrm{t}}=\frac{600 \times 10 \times 25}{60}=2500 \mathrm{watt}\)

K.E. of water per second = \(\frac{1}{2} \frac{\mathrm{mv}^2}{\mathrm{t}}\)

= \(\frac{1}{2} \frac{600 \times 25 \times 25}{60}=\frac{6250}{2}\)

= 3125 watts.

∴ Power of motor P = 2500 + 3125 = 5625 watt = 5.625 K.W.

Question 5. A block of mass 5 kg initially at rest at the origin is acted on by a force along the X-positive direction represented by F = (20 + 5x)N Calculate the work done by the force during the displacement of the block from x = 0 to x = 4m.
Solution:

Mass of block, m = 5 kg

Force acting on the block, F = (20 + 5x) N

If ‘w’ is the total amount of work done to displace the block from x = 0 to x = 4m then

W = \(\int \mathrm{dW}=\int_{x=0}^{x=4} F d x=\int_{x=0}^{x=4}(20+5 x) d x\)

= \(\left[20 x+\frac{5 x^2}{2}\right]_0^4=\left[20 \times 4-0+\frac{5}{2} \times(4)^2-0\right]\)

⇒ W=80+40=120 J

Question 6. A block of mass 5 kg is sliding down a smooth inclined plane as shown. The spring arranged near the bottom of the inclined plane has a force constant of 600 N/m. Find the compression in the spring at the moment the velocity of the block is maximum.

Solution:

Mass of the block, m = 5kg

Force constant, K = 600 N m_^-1.

From sin = \(\frac{3}{5}\)

The force produced by the motion in the block,

F = mg sin θ

⇒ F = 5 x 9.8 x \(\frac{3}{5}\) = 29.4 N

But force constant K = \(\frac{F}{x}\)

∴ x = \(\frac{\mathrm{F}}{\mathrm{K}}=\frac{29.4}{600}=0.05 \mathrm{~m}=5 \mathrm{~cm}\)

Question 7. A force F = –\(\frac{K}{x^2}\) (x ≠ 0) nets on a particle along the X-axis. Find the work done by the force in displacing the particle from x = + a to x = + 2a. Take K as a positive constant.
Solution:

Force acting on the particle, F = –\(\frac{K}{x^2}\)

The total amount of work done to displace the particle from x = + a to x = + 2a is,

W = \(\int d W=\int_{x=a}^{x=2 a} F d x\)

⇒ W = \(\int_{x=a}^{2 a} \frac{-K}{x^2} d x=-K \int_{x=a}^{x=2 a} \frac{1}{x^2} d x\)

= \(-K \int_{x=a}^{x=2 a} x^{-2} d x=K\left[x^{-1}\right]_{x=a}^{x=2 a}\)

= \(K\left[\frac{1}{2 a}-\frac{1}{a^{\prime}}\right]=K\left[\frac{-1}{2 a}\right]=\frac{-K}{2 a}\)

Question 8. A force F acting on a particle varies with the position x as shown in the graph. Find the work done by the force in displacing the particle from x = -a to x = + 2a.

Solution:

Average force acting on the particle, \(\mathrm{F}=\frac{-\mathrm{b}+2 \mathrm{~b}}{2}=\frac{\mathrm{b}}{2}\)

The amount of work done by the force to displace the particle from x = -a to x = +2a is,

W = \(\int F d x \Rightarrow W=\int_{x=-a}^{x=2 a} F d x=\int_{x=-a}^{x=2 a} \frac{b}{2} d x\)

= \(\frac{b}{2} \int_{x=-a}^{x-2 a} d x=\frac{b}{2}[x]_{x=-a}^{x=2 a}=\frac{b}{2}[2 a+a]=\frac{3 a b}{2}\)

Question 9. From a height of 20 in above a horizontal floor, a ball is thrown down with an initial velocity of 20 m/s. After striking the floor, the ball bounces to the same height from which it was thrown. Find the coefficient of restitution for the collision between the ball and the floor. (g = 10 m/s²)
Solution:

Initial velocity = u¹ = 20 m/s, h = 20 m, g = 10 m/s²

Velocity of approach, \(u^2=u^{1^2}=u^{1^2}+2 a s=400+2 \times 10 \times 20\)

⇒ \(u^2=400+400=800 \Rightarrow u=20 \sqrt{2}\)

Height of rebounce = h = 20 m.

∴ Velocity of separation \(v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 20}=\sqrt{400}=20\)

Coefficient of restitution, \(\mathrm{e}=\frac{\text { Velocity of separation }}{\text { Velocity of approach }}=\frac{20}{20 \sqrt{2}}=\frac{1}{\sqrt{2}}\)

Work, Energy, And Power Questions And Answers KSEEB Physics 

Question 10. A ball falls from a height of 10 m onto a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is \(\frac{1}{\sqrt{2}}\), then what is the total distance travelled by the ball before it ceases to rebound?
Solution:

Height from which the ball is allowed to fall, h = 10 m

Coefficient of restitution between the hard horizontal floor and the ball, e = \(\frac{1}{\sqrt{2}}\)

∴ Total distance travelled by the ball before it ceases to rebound, \(\mathrm{d}=\mathrm{h}\left[\frac{1+\mathrm{e}^2}{1-\mathrm{e}^2}\right]=10\left[\frac{1+(1 / \sqrt{2})^2}{1-(1 / \sqrt{2})^2}\right]=30 \mathrm{r}\)

Question 11. In a ballistics demonstration, a police officer fires a bullet of mass 50g with a speed of 200 ms^-1 on soft plywood of thickness 2 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet?
Answer:

Mass of bullet m = 50g = 0.05 kg

Initial Velocity V₀ = 200 m/s

Initial K.E. = \(\frac{1}{2} \mathrm{mv}_0{ }^2=\frac{1}{2} \times \frac{50}{1000} \times 200 \times 200\)

Final K.E. = 10% of \(1000 \mathrm{~J}=100 \mathrm{~J}\)

But K.E. = \(\frac{1}{2} \mathrm{mv}_1{ }^2 \Rightarrow \mathrm{v}_1=\sqrt{\frac{2 \mathrm{KE}}{\mathrm{m}}}\)

∴ \(\mathrm{v}_1=\sqrt{\frac{2 \times 100}{0.05}}=\sqrt{4000}=20 \sqrt{10}=63.2 \mathrm{~m} / \mathrm{s}\)

Question 12. Find the total energy of a body of 5 kg mass, which is at a height of 10 m from the earth and falling downwards straightly with a velocity of 20 m/s. (Take the acceleration due to gravity as 10 m/s²)
Answer:

Mass m = 5 kg;

Height h = 10 m; g = 10 m/s²

Velocity v = 20 m/s.

Total energy = PE + KE = mgh + \(\frac{1}{2}\) mv²

= \((5 \times 10 \times 10)+\frac{1}{\not 2} \times 5 \times 20^{10} \times 20\)

= \(500+1000=1500 \mathrm{~J}\)

KSEEB Class 11 Physics Solutions For Chapter 5 Motion Very Short Type Questions And Answers

Question 1. Why are spokes provided in a bicycle wheel?
Answer:

The spokes of the cycle wheel increase its moment of inertia. The greater the moment of inertia, the greater the opposition to any change in uniform rotational motion.

As a result, the cycle runs smoother and steadier. If the cycle wheel had no spokes, the cycle would be driven with jerks and hence unsafe.

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. What is inertia? What gives the measure of inertia?
Answer:

The inability of a body to change its state by itself is known as inertia. The mass of a body is a measure of its Inertia.

Types Of Inertia

1. Inertia of rest
2. Inertia of motion
3. Inertia of direction.

Question 3. According to Newton’s third law, every force is accompanied by an equal and opposite force. How can a movement ever take place?
Answer:

From Newton’s third law action = -reaction. But action and reaction are not working on the same system. So they will not cancel each other. Hence, motion is possible.

Question 4. When a bullet is fired from a gun, the gun gives a kick in the backward direction. Explain.
Answer:

The firing of a gun is due to Internal form. Terminal forces do not change The momentum of the system. Before firing m₁u₁ + m₂u₂ = 0.

Since the system is at rest after firing m₁v₁ + m₂v₂ = 0(or) m₁v₁ = – m₂v₂. So gun and bullet will move in opposite directions to satisfy the law of conservation of linear momentum.

Question 5. Why does a heavy rifle not recoil as strongly as a light rifle using the same cartridges?
Answer:

Velocity (or) recoil \(v=\frac{m v}{M}\) i.e., the ratio of momentum of the bullet to the mass of the gun. If the mass of the gun is high then the velocity of the recoil is less with the same cartridge.

KSEEB Physics Class 11 Laws Of Motion Notes

Question 6. If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain.
Answer:

Explosion is due to internal forces. From the law of conservation of linear momentum, internal forces cannot change the momentum of the system. So after explosion m₁v₁ + m₂v₂ = 0 or m₁v₁ = – m₂v₂. According to the law of conservation of linear momentum, they will fly in opposite directions.

Question 7. Define force. What are the basic forces in nature? Answer:

Force is that which changes (or) tries to change the state of a body.

The Basic Forces In Nature Are:

1. Gravitational forces,
2. Electromagnetic forces,
3. Nuclear forces.

Question 8. Can the coefficient of friction be greater than one?
Answer:

Yes. Generally coefficient of friction between the surfaces is always less than one. But under some special conditions like on extremely rough surfaces coefficient of friction may be greater than one.

Question 9. Why does the car with a flat tire stop sooner than the one with inflated tires?
Answer:

Due to the flatting of tires, the frictional force increases. Because rolling frictional force between the surfaces Is proportional to the area of contact. Area of contact Increases for flattened tires. So rolling frictional force Increases and the car will be stopped quickly.

Question 10. A horse has to pull harder during the start of the motion than later. Explain.
Answer:

To start motion in a body we must apply force to overcome static friction(F_s = μ_smg). When once motion is started between the bodies then kinetic frictional force comes into act.

Kinetic friction (F_k = μ_kmg) is always less than static friction. So it is tougher to start a body from rest than to keep it in. motion.

Laws Of Motion Question And Answers In KSEEB Physics

Question 11. What happens to the coefficient of friction if the weight of the body is doubled?
Answer:

When the weight of the body is doubled still then there is no change in coefficient of friction. Because frictional force oc normal reaction.

So when the weight of a body is doubled then frictional force and normal reaction will also become doubled and the coefficient of friction remains constant.

Laws Of Motion Solutions KSEEB Class 11 Physics Short Type Questions And Answers

Question 1. A stone of mass 0.1 kg is thrown vertically upwards. Give the magnitude and direction of the net force on the stone

1. During its upward motion,
2. During its downward motion,
3. At the highest point, where it momentarily comes to rest.

Answer:

Mass of stone, m = 0. 1 kg.

1. During upward motion force acts downwards due to acceleration due to gravity.

Magnitude of force F = mg = 0.1 x 9.8 = 0.98 N(↓)

During downward motion force acts downward.

Magnitude of force F = mg = 0.1 x 9.8 = 0.98N(↓)

3. At the highest point velocity v = 0. Hut still g will act on it only In downward motion so resultant forceF = 0.98 N. downward.

Note: In the entire Journey of the body force due to gravitational pull acts only In a downward direction.

4. If the body is thrown with an angle of 30° with the horizontal then the vertical component of gravitational force does not change, hence in this case downward force F = mg = 0.1 x 9.8 = 0. 98 newton.

Question 2. Define the terms momentum and impulse. , State and explain the law of conservation of linear momentum. Give examples.
Answer:

Momentum \((\overline{\mathbf{p}})\): It is the product of mass and velocity of a body.

Momentum \((\overline{\mathbf{p}})\) = mass(m) x velocity(v)

∴ \((\overline{\mathbf{p}})\) = m \((\overline{\mathbf{v}})\)

Impulse (J): When a large force(F) acts on a body for a small time(t) then the product of force and time is called Impulse.

Impulse(J) = Force(F) x time(t)

∴ Impulse(J) = F x t

KSEEB Class 11 Physics Laws of Motion 

Law Of Conservation Of Linear Momentum: There is no external force acting on the system. The total linear momentum of the isolated system remains constant.

Proof: Let two bodies of masses say A and B are moving with initial momenta PA and PB collided with each other.

During collision they are in contact for a small time say Δt. During this time of contact, they will exchange their momenta. Let the final momenta of the bodies be P_A¹ and P_B¹. Let force applied by A on B is F_AB and force applied by B on A is F_BA.

From Newton’s 3rd Law F_AB = F_BA or F_AB Δt = F_BA Δt

From 2nd Law \(F_{A B} \Delta t=P_A^1-P_A\) change in momentum of A.

\(F_{B A} \Delta t=P_{11}^{\prime}-P_B\) change in momentum of B,

i.e., the sum of momentum before collision Is equal to the sum of momentum after collision.

Question 3. Why are shock absorbers used in motorcycles and cars?
Answer:

1. When vehicles are passing over the vertices and depressions of a rough road they will collide with them for a very short period. This causes an impulse effect.
2. Due to the large mass and high speed of the vehicles, the magnitude of impulse is also high. Impulse may cause damage to the car or even to the passengers in it. \(\mathrm{F} \propto \frac{1}{\Delta \mathrm{t}}\)
3. The bad effects of impulse are less if the time of contact is longer. Impulse J = F.t. For the same magnitude of impulse(change in momentum) if the time of contact is high force acting on the vehicle is less.

Shock absorbers will absorb the impulse and release the same force slowly. This is due to the large time constant of the springs.

So shock absorbers are used in vehicles to reduce impulse effects.

Question 4. Explain the terms limiting friction, dynamic friction, and rolling friction.
Answer:

Limiting Friction: Frictional forces always oppose relative motion between the bodies. These forces are self-adjusting forces. Their magnitude will increase to some extent with the value of applied force.

The maximum frictional force between the bodies at rest is called”limiting friction”.

Dynamic (Or) Kinetic Friction: When the applied force is equal to or greater than limiting friction then the body will move. When once motion is started then frictional force will abruptly fall to a minimum value.

The frictional force between moving bodies Is called dynamic (or) kinetic friction. Kinetic friction Is always less than limiting friction.

Rolling Friction: The resistance encountered by a rolling body on a surface is called rolling friction.

Practice Questions For Laws Of Motion KSEEB Physics

Question 5. Explain the advantages and disadvantages of friction.
Answer:

Advantages Of Friction:

1. We are able to walk because of friction.
2. It is impossible for a car to move on a slippery road.
3. The braking system of vehicles works with the help of friction.
4. Friction between roads and tires provides the necessary external force to accelerate the car.
5. Transmission of power to various parts of a machine through belts is possible by friction.

Disadvantages Of Friction:

1. In many cases, we will try to reduce friction because it dissipates energy into heat.
2. It causes wear and tear to machine parts which causes frequent replacement of machine parts.

Question 6. Explain Friction. Mention the methods used to decrease friction.
Answer:

Friction: It is a contact force parallel to the surfaces in contact. Friction will always oppose relative motion between the bodies.

Methods To Reduce Friction:

1. Polishing: Friction is caused due to surface irregularities. So by polishing friction can be reduced to some extent.
2. Lubricants: By using lubricants friction can be reduced. Lubricants will spread as an ultra-thin layer between the surfaces in contact and friction decreases.
3. Stream Elining: By making a front portion of vehicles in a curved shape friction due to air can be reduced.
4. Ball Bearings: Ball bearings are used to reduce friction between machine parts. Ball bearing will convert sliding motion into rolling motion. As a result, friction is reduced.

Question 7. State the laws of rolling friction.
Answer:

When a body is wiling over the other, then friction between the bodies is known as rolling friction.

Rolling friction coefficient, \(\mu_{\mathrm{r}}=\frac{\text { Rolling friction }}{\text { Normal reaction }}=\frac{\mathrm{Fr}}{\mathrm{N} \cdot \mathrm{R}}\)

Laws Of Rolling Friction:

1. Rolling friction will develop a point of contact between the surface and the rolling sphere. For objects like wheels line of contact will develop.
2. Rolling friction(f_r) has the least value for a given normal reaction when compared with static friction(f_s) or kinetic friction (f_k).
3. Rolling friction is directly proportional to F = ma from it. to normal reaction, f_r ∝ N.
4. In rolling friction, the surfaces in contact will get momentarily deformed a little.
5. Rolling friction depends on the area of contact. Due to this reason, friction increases when air pressure is less in tires (Flattened tires).
6. Rolling friction is inversely proportional to the radius of the rolling body \(\mu_r \propto \frac{1}{r}\).

Practice Questions For Laws Of Motion KSEEB Physics

Question 8. Why is pulling the lawn roller preferred to pushing it?
Answer:

Let a lawn roller be pulled by means of a force F with some angle θ to the horizontal. By resolving the force into two components.

1. The horizontal component F cos θ is useful. to pull the body.
2. The vertical component F sin θ opposes the weight

So N.R. = mg – F sin θ

But frictional force μ N.R.

∴ Frictional force [μ(mg- F sin θ)] decreases,

So it is easier to pull the body.

When the lawn roller is pushed by a force, the vertical component F sin θ causes the apparent increase in the weight of the object.

So the normal reaction N.R. = mg + F sin θ

∴ Frictional force [μ(mg + F sinθ)] increases and it will be difficult to pull the body.

KSEEB Class 11 Physics Chapter 5 Laws Of Motion Long Type Questions And Answers

Question 1.

1. State Newton’s second law of motion. Hence, derive the equation of motion F = ma from it.
2. A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body?

Answer:

1. Newton’s 2nd law: The rate of change of momentum of a body is proportional to external force and acts along the direction of force applied.

i.e., \(\frac{d p}{d t} \propto F\)

Derivation Of Equation F = ma: According to Newton’s 2nd law.

We know \(\frac{\mathrm{dp}}{\mathrm{dt}} \propto \vec{\mathrm{F}} \Rightarrow \mathrm{F}=\mathrm{k} \cdot \frac{\mathrm{d} \overline{\mathrm{p}}}{\mathrm{dt}}\)…..(1)

But  \(\vec{p}\) = momentum of the body = m \(\bar{v}\)

∴ F = \(\mathrm{k} \cdot \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{m} \overline{\mathrm{v}}) \Rightarrow \mathrm{km} \frac{\mathrm{d} \overline{\mathrm{v}}}{\mathrm{dt}}\)……(2)

But \(\frac{\mathrm{d} \overline{\mathrm{v}}}{\mathrm{dt}}\) = Rate of change in velocity = acceleration ‘a’.

∴ F = ma…… (3)

Here, k = constant.

The proportional constant is made equal to one, by properly selecting the unit of force.

∴ F = ma.

2. Force On A Body Moving In A Circular Path: Let a body of mass ‘m’ be moving in a circular path of radius ‘r’ with constant speed.

The velocity of the body is given by the tangent drawn at that point. Since velocity is changing continuously the body will have acceleration.

So the body will experience some acceleration. This is called normal acceleration (or) centripetal acceleration.

Question 2. Define the Angle of friction and Angle of repose. Show that the angle of friction is equal to the angle of repose for a rough inclined plane.

A block of mass 4 kg is resting on a rough horizontal plane and is about to move when a horizontal force of 30 N is applied to it. If g = 10 m/s². Find the total contact force exerted by the plane on the block.

Answer:

Angle Of Friction: The angle made by the resultant of the Normal reaction and the limiting friction with the Normal reaction is called the angle of friction (ø).

Angle Of Repose: Let a body of mass m is placed on a rough inclined plane. Let the angle with the horizontal ‘θ’ be gradually increased then for a particular angle of inclination (say a) the body will just slide down without acceleration.

This angle θ = α is called the angle of repose. At this stage, the forces acting on the body are in equilibrium

Equation For Angle Of Repose: Force acting on the body in a vertically downward direction = W = mg.

By resolving this force into two components.

1. Force acting along the inclined plane in a downward direction = mg sinθ.
   • This component is responsible for downward motion.
2. The component mg cos θ. which is balanced by the normal reaction.

If the body slides down without acceleration resultant force on the body is zero, then

mg sin θ = Frictional force (f_k)

mg cos θ = Normal reaction (N.R.)

But coefficient friction \(\mu_{\mathrm{k}}=\frac{\mathrm{f}_{\mathrm{k}}}{\mathrm{N} \cdot \mathrm{R}}=\frac{\mathrm{mg} \sin \theta}{\mathrm{mg} \cos \theta}=\tan \theta\)

Hence θ = α is called the angle of repose.

∴ μ_k = tan α

Hence tangent of the angle of repose (tan θ) is equal to the coefficient of friction (f_k) between the bodies.

2. Given m = 4 kg

g = 10 m/s²

Normal reaction N = mg

N = 4 x 10 = 40 N

Horizontal frictional force f = 30 N

Total contact force \(\mathrm{F}=\sqrt{\mathrm{f}^2+\mathrm{N}^2}\)

F = \(\sqrt{30^2+40^2}\)

F = \(\sqrt{900+1600}\)

F = \(\sqrt{2500}\)

∴ F = 50 N

Chapter 5 Solutions For Laws Of Motion KSEEB Physics Problems

Question 1. The linear momentum of a particle as a function of time is given by, p = a + bt, where a and b are positive constants. What is the force acting on the particle?
Solution:

Linear momentum of a particle, p = a + bt

We know that force acting on a particle is equal to the rate of change of linear momentum.

i.e., \(\vec{F}=\frac{d p}{d t} \Rightarrow \vec{F}=\frac{d}{d t}(a+b t)\)

⇒ \(\vec{F}=\frac{d}{d t}(b t)=b \frac{d}{d t}=b \times 1=b\)

Question 2. Calculate the time needed for a net force of 5 N to change the velocity of a 10 kg mass by 2 m/s.
Solution:

Force, F = 5N

Change in velocity, v – u = 2ms^-1

Mass, m = 10 kg

From Newton’s second law of motion,

F = \(m a=m \frac{(v-u)}{t}\)

∴ t = \(\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{F}}=\frac{10 \times 2}{5}=4 \mathrm{~s}\)

Question 3. A ball of mass ‘m’ is thrown vertically upward from the ground and reaches a height ‘h’ before momentarily coming to rest, If ‘g’ is the acceleration due to gravity. What is the impulse received by the ball due to gravity force during its flight? (neglect air resistance)
Solution:

Impulse, J = force x time

⇒ J = \(\mathrm{ma} \times \frac{2 \mathrm{u}}{\mathrm{g}}\);

Here, a = g and \(\mathrm{u}=\sqrt{2 \mathrm{gh}}\)

Now \(\mathrm{J}=\mathrm{mg} \times \frac{2 \times \sqrt{2 \mathrm{gh}}}{\mathrm{g}} \Rightarrow \mathrm{J}=2 \mathrm{~m} \sqrt{2 \mathrm{gh}}\)

∴ Impulse \(\mathrm{J}=\sqrt{8 \mathrm{~m}^2 \mathrm{gh}}\)

KSEEB Physics Laws of Motion Questions And Answers

Question 4. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^-1 to 3.5 m s^-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:

Mass of the body, m = 3.0 kg

The initial velocity of the body, u = 2.0 ms^-1

The final velocity of the body, v = 3.5 ms^-1

Time, t = 25 s

From Newton’s second law of motion,

F = \(\mathrm{ma}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}=\frac{3.0(3.5-2.0)}{25}=0.18 \mathrm{~N}\)

∴ Magnitude of force acting on the body, F = 0.18 N. The direction of force acting on the body is along the direction of motion of the body because force is positive.

Question 5. A man in a lift feels an apparent weight W when the lift is moving up with a uniform acceleration of 1/3rd of the acceleration due to gravity. If the same man was in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is his| apparent weight?
Solution:

Case (1): When Lift Is Moving Upwards:

The apparent weight of the man = W

Acceleration, a = g/3

The apparent weight of the man, when the lift is moving upwards, is,

W = \(\mathrm{m}(\mathrm{a}+\mathrm{g})=\mathrm{m}\left(\frac{\mathrm{g}}{3}+\mathrm{g}\right) \)

= \(\frac{4}{3} \mathrm{mg}=\frac{4}{3} \mathrm{~N}\) (because \(\mathrm{N}=\mathrm{mg}\))

∴ \(\mathrm{N}=\frac{3}{4} \mathrm{~W}\)

Case (2): When the lift is moving downwards: Let W’ be (lie apparent weight of the man Acceleration, a = g/2

The apparent weight of the man when the lift is moving downwards is, W” = m(g – a) = m (g – g/2)

= \(\frac{1}{2}\) mg = \(\frac{N}{2}\) [N = mg]

= \(\frac{1}{2}\)(\(\frac{3}{4}\) W) = \(\frac{3}{8}\) W

Question 6. A container of mass 200 kg rests on the back of an open truck. If the truck accelerates at 1.5 m/s², what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck?
Solution:

Mass of the container, m = 200 kg

Acceleration of truck, a = 1.5 ms^-2

Coefficient of static friction, u_s = \(\frac{a}{g}\)

= \(\frac{1.5}{9.8}\) = 0.153

Question 7. A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically downwards with an initial speed of 10 m/s. If acceleration due to gravity is 10m/s², what is the separation between the fragments 2s after the explosion?
Solution:

Case (1): (for a downward-moving fragment)

Initial velocity, u = 10 ms^-1

Acceleration, a = +g = 10 ms^-1

Time, t = 2s

From the equation of motion, s = ut + \(\frac{1}{2}\) at² the distance moved in downward direction is,

s₁ = 10 x 2 + \(\frac{1}{2}\) x 10 x (2)² = 40

Case (2) (For Upward Moving Fragment) Given that two fragments are identical hence, after explosion, the fragments move in opposite directions.

Here the first fragment moves in a downward direction, hence, the second fragment moves upward direction.

Again from s = ut + \(\frac{1}{2}\) at² we can write,

s₂ = – 10 x 2+ \(\frac{1}{2}\) x 10 x 4 =-20 + 20 = 0 m

Separation between the fragments 2s after the explosion = s₁ – s₂ = 40 – 0 = 40m

Question 8. A fixed pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3 kg on the other side. Another 3 kg is hung from the other 3 kg as shown with another light string. If the system is released horn rest, find the common acceleration. (g = 10 m/s²)

Solution:

Here, m₁ = 3 + 3 = 6 kg; m₂ = 4 kg; g = 10 ms^-2

Acceleration of the system, \(a=\left(\frac{m_1-m_2}{m_1+m_2}\right) g \Rightarrow a=\left(\frac{6-4}{6+4}\right) \times 10=2 \mathrm{~ms}^{-2}\)

Question 9. A block of mass of 2 kg slides on an inclined plane that makes an angle of 30° with the horizontal. The coefficient of friction between the block and the surface is √3/2

1. What force should be applied to the block so that it moves down without any acceleration?
2. What force should be applied to the block so that it moves up without any acceleration?

Solution:

Mass of the block, m = 2kg

The angle of Inclination, θ = 30°

Coefficient of friction between the block and the surface, \(\mu=\frac{\sqrt{3}}{2}\)

1. The required force to move the block down without acceleration is,

F = \(\mathrm{mg}\left(\sin \theta-\mu_{\mathrm{k}} \cos \theta\right)\)

= \(2 \times 9.8\left(\sin 30^{\circ}-\frac{\sqrt{3}}{2} \times \cos 30^{\circ}\right)\)

= \(2 \times 9.8\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right)=4.9 \mathrm{~N}\)

2. The required force to move the block up without any acceleration is,

F = \(\mathrm{mg}\left(\sin \theta+\mu_{\mathrm{k}} \cos \theta\right)\)

= \(2 \times 9.8\left(\sin 30^{\circ}+\frac{\sqrt{3}}{2} \cos 30^{\circ}\right)\)

= \(2 \times 9.8\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right)=24.5 \mathrm{~N}\)

KSEEB Class 11 Physics Chapter 5 Laws Of Motion 

Question 10. A block is placed on a ramp of parabolic shape given by the equation y = x²/20, see Figure.

If μ_s = 0.5, what is the maximum height I above the ground at which the block can be placed without slipping? \(\left(\tan \theta=\mu_{\mathrm{s}}=\frac{d y}{d x}\right)\)

Solution:

For the body not to drop

mg cos θ = μ mg sin θ

⇒ tan θ = μ given μ = 0.5
.
But tan θ = \(\frac{dy}{dx}\) slope of parabolic region

⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mu=0.5\)…..(1)

given \(\mathrm{y}=\frac{\mathrm{x}^2}{20}\)

∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{x}}{20}=\cdot \frac{\mathrm{x}}{10}\)…..(2)

∴ slope \(\frac{\mathrm{y}}{\mathrm{x}}=0.5 \Rightarrow 0.5=\frac{\mathrm{x}}{10} \Rightarrow \mathrm{x}=5\)…..(3)

From eq. (1) and (3) vertical height, y = \(\frac{x^2}{20}=\frac{5 \times 5}{20}=1.25 \mathrm{~m}\)

Question 11. A block of metal of mass 2 kg on a horizo¬ntal table is attached to a mass of 0.45 kg by a light string passing over a frictionless pulley at the edge of the table. The block is subjected to a horizontal force by allowing the 0.45 kg mass to fall. The coefficient of sliding friction between the block and the table is 0.2. Calculate

1. The initial acceleration,
2. The tension in the string,
3. The distance the block would continue to move if, after 2 s of motion, the string should break.

Solution:

Mass of the first block, m₁ = 0.45 kg

Mass of the second block, m₂ = 2kg

coefficient of sliding friction between the block and table, μ = 0.2

1. Initial acceleration, \(a=\left(\frac{m_1-\mu m_2}{m_1+m_2}\right) g\)

⇒ a = \(\left(\frac{0.45-0.2 \times 2}{0.45+2}\right) \times 9.8\)

⇒ a = \(0.2 \mathrm{~ms}^{-2}\)

2. Tension in the string

T = \(\frac{\left(\mathrm{m}_1+\mu \mathrm{m}_2\right) g+\left(\mathrm{m}_2-\mathrm{m}_1\right) \mathrm{a}}{2}\)

T = \(\frac{(0.45+0.2 \times 2) 9.8+(2-0.45) 0.2}{2}\)

= \(\frac{(0.85 \times 9.8)+(1.55 \times 0.2)}{2}\)

T = \(\frac{8.33+0.31}{2}=\frac{8.64}{2}=4.32\) Newton

3. Velocity of string after 2 sec=u in this case; \(\mathrm{u}^{\prime}=0\)

∴ \(\mathrm{u}=\mathrm{u}^{\prime}+\mathrm{at}=0+0.2 \times 2=0.4 \mathrm{~m} / \mathrm{s}\)

Stopping distance, \(\mathrm{s}=\frac{\mathrm{u}^2}{2 \mu \mathrm{g}}\)

= \(\frac{0.4 \times 0.4}{2 \times 0.2 \times 9.8}=\frac{0.4}{9.8}=0.0408 \mathrm{~m}\)

or \(\mathrm{s} \simeq 4.1 \mathrm{~cm}\)

KSEEB Class 11 Physics Solutions For Chapter 5 Laws Of Motion

Question 12. On a smooth horizontal surface a block A of mass 10 kg is kept. On this block, a second block B of mass 5 kg is kept. The coefficient of friction between the two blocks is 0.4. A horizontal force of 30 N is applied on the lower block as shown. The force of friction between the blocks is (take g = 10 m/s²)

Solution:

The mass of block ‘A’ is m_A = 10 kg

The mass of block ‘B’ is m_B = 5 kg

Applied horizontal force, F = 30 N

Coefficient of friction between two blocks,  μ = 0.4

The frictional force of block ’B’  Is f = μ mg

⇒ f = 0,4 x 5 x 10 = 20N

∴ The frictional force acting between the two blocks, = F – f = 30 – 20 = 10N

Question 13. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 ms^-1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball).
Solution:

Impulse = change in momentum = (0.15 x 12)-(-0.15 x 12) = 3.6 NS .

In the direction from the batsman to the bowler.

Question 14. A force 2\(\bar{i}\) + \(\bar{j}\) — \(\bar{k}\) Newton acts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is 4\(\bar{i}\) + 2\(\bar{j}\) – 2\(\bar{k}\) m/s. What is the mass of the body?
Solution:

Force F = 2\(\bar{i}\) + \(\bar{j}\) – \(\bar{k}\) = 20sec

Initial velocity u₀ = 0.

Final velocity U = 4\(\bar{i}\) + 2\(\bar{j}\) – 2\(\bar{k}\)

Mass of the body m = \(\frac{F}{a}\)

But acceleration \(a=\frac{U-U_0}{t}\)

∴ \(\mathrm{m}=\frac{\text { F.t }}{\mathrm{U}-\mathrm{U}_0}=\frac{20 \cdot|\mathrm{F}|}{\left|\mathrm{U}-\mathrm{U}_0\right|}\)

= \(\frac{20 \sqrt{2^2+1+1}}{\sqrt{4^2+2^2+2^2}-0}=\frac{20 \sqrt{6}}{\sqrt{24}}\)

∴ \(\mathrm{m}=\frac{20 \sqrt{6}}{\sqrt{4 \times 6}}=\frac{20 \sqrt{6}}{2 \sqrt{6}}=10 \mathrm{~kg} \text {. }\)

Question 15. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
Solution:

Here, F = – 50N, m = 20 kg

u = 15 ms^-1, v = 0, t = ?

From F = \(m a, a=\frac{F}{m}=\frac{-50}{20}=-2.5 \mathrm{~ms}^{-2}\)

From \(\mathrm{v}=\mathrm{u}+\mathrm{at} ; 0=15-2.5 \mathrm{t}\)

t = \(\frac{15}{2.5}=6 \mathrm{~s}\)

Question 16. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6N. Give the magnitude and direction of the acceleration of the body.
Solution:

Here, m = 5 kg, \(\vec{a}\) = ?

⇒ \(\vec{F}_1=\overrightarrow{O A}=8 \mathrm{~N} ; \vec{F}_2=\overrightarrow{O B}=6 \mathrm{~N}\)

Resultant force \(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{OC}}=\sqrt{\mathrm{F}_1^2+\mathrm{F}_2^2}\)

= \(\sqrt{8^2+6^2}=10 \mathrm{~N}\)

If \(\angle A O C=\theta, \tan \theta=\frac{\mathrm{AC}}{\mathrm{OA}}=\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{6}{8}=0.75\)

∴ \(\theta=36^{\circ} 52^{\prime}\)

This is the direction of the resultant force and hence the direction of acceleration of the body as shown.

Also, \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{10}{5}=2 \mathrm{~ms}^{-2}\)

Laws Of Motion Solutions KSEEB Class 11 Physics

Question 17. The driver of a three-wheeler moving with a speed of 36 km /h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 seconds just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Solution:

Here, u = 36 km/h = 10 m/s, v = 0, t = 4s

m = 400 + 65 = 465 kg

Retarding force = \(\mathrm{F}=\mathrm{ma}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\)

= \(\frac{465(0-10)}{4}=-1162.5 \mathrm{~N}\)

Question 18. A rocket with a lift-off mass of 20,000 kg Is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Solution:

Here, m = 20000 kg = 2 x 10⁴ kg

Initial acceleration, a = 5 ms^-2;

Thrust, F =?

Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms^-2.

Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 ms^-2.

As thrust = force = mass x acceleration

∴ F = 2 X 10⁴ X 14.8 = 2.96 X 10⁵  N

Question 19. A man of mass 70 kg stands on a weighing scale in a lift which is moving

1. Upwards with a uniform speed of 10 ms^-1,
2. Downwards with a uniform acceleration of 5 ms^-2,
3. Upwards with a uniform acceleration of 5 m s^-2, What would be the readings on the scale in each case?
4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Solution:

Here, m = 70 kg, g = 10 m/s²

The weighing machine in each case measures the reaction R i.e., the apparent weight.

1. When the lift moves upwards with a uniform speed, its acceleration is zero.

∴  R = mg = 70 x 10 = 700 N

2. When the lift moves downwards with a = 5 ms^-2

∴ R = m(g- a) = 70 (10 – 5) = 350 N

3. When the lift moves upwards with a = 5 ms^-2

∴ R = m (g + a) = 70 (10 + 5) = 1050 N

4. If the lift were to come down freely under gravity, downward acceleration. a = g

∴ R = m (g – a) = m (g – g) = Zero.

Question 20. Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface arc tied to the ends of a light string, a horizontal force F = 000 N Is applied to

1. A,
2. B along the direction of hiring.

What is the tension In the string in each ease?

Solution:

Here, F = 600 N m₁ = 10 kg, m₂ = 20 kg

Let T be the tension in the string and a be the acceleration of the system, in the direction of force applied.

∴ a = \(\frac{F}{m_1+m_2}=\frac{600}{10+20}=20 \mathrm{~m} / \mathrm{s}^2\).

1. When force is applied on lighter block A, T = m₂a = 20 x 20 N = 400 N
2. When force is applied on heavier block B, T = m₁a = 10 x 20 N T = 200 N

Which is different from the value of T in case (1).

Hence our answer depends on which mass end, the force is applied.

KSEEB Class 11 Physics Chapter 5 Laws Of Motion

Question 21. Two masses 8 kg and 12 kg are connected at the two ends of a light in an extensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Solution:

Here, m₂ = 8 kg, ; m₁ = 12kg

As a = \(\frac{\left(m_1-m_2\right) g}{m_1+m_2}=\frac{(12-8) 9.8}{12+8}=\frac{39.2}{20}\) = \(1.96 \mathrm{~ms}^{-2}\)

Again, \(T=\frac{2 m_1 m_2 g}{m_1+m_2}=\frac{2 \times 12 \times 8 \times 9.8}{(12+8)}\)

= 94.1 N

Question 22. A nucleus is at rest In the laboratory frame of reference. Show that If It disintegrates Into two smaller nuclei the products must move in opposite directions.
Solution:

Let m₁, m₂ be the masses of products and \(\overrightarrow{v_1}, \overrightarrow{v_2}\) be their respective velocities. Therfore, total linear momentum after disinte-gration = \(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\).

Before disintegration, the nucleus is at rest. Therefore, its linear momentum before disintegration is zero.

According to the principle of conservation of linear momentum, \(m_1 \overline{v_1}+m_2 \overline{v_2}=0 \text { or } \overline{v_2}=\frac{-m_1 \overline{v_1}}{m_2}\)

Negative sign shows that \(\overrightarrow{\mathrm{v}_1} \text { and } \overrightarrow{\mathrm{v}_2}\) are in opposite directions.

Question 23. Two billiard balls each of mass 0.05 kg moving in opposite directions with a speed of 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Solution:

Here, initial momentum of the ball A = 0.05 (6) = 0.3 kg ms^-1

As the speed is reversed on collision, the final momentum of the ball A = 0.05 (-6) = – 0.3 kg ms^-1

Impulse imparted to ball A = change in momentum of ball A = final momentum – initial momentum = – 0.3 – 0.3 = – 0.6 kg ms^-1.

Question 24. A shell of mass 0.02 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun?
Solution:

Here, the mass of the shell, m = 0.02 kg.

mass of gun, M = 100 kg

muzzle speed of shell, V = 80 ms^-1

recoil speed of gun, v =?

According to the principle of conservation of linear momentum, mV + Mν = 0

or, \(\mathrm{v}=-\frac{\mathrm{mV}}{\mathrm{M}}=\frac{-0.02 \times 80}{100}=0.016 \mathrm{~ms}^{-1}\)

Solutions For Laws of Motion KSEEB Physics Chapter 5 

Question 25. A slone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 ni with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Solution:

Here, m = 0.25 kg, r = 1.5 m;

n = \(40 \mathrm{rpm}=\frac{40}{60} \mathrm{rps}=\frac{2}{3} \mathrm{rps}, \mathrm{T}=?\)

T = \(\mathrm{mr} \omega^2=\mathrm{mr}(2 \pi \mathrm{n})^2=4 \pi^2 \mathrm{mr} \mathrm{r}^2\)

T = \(4 \times \frac{22}{7} \times \frac{22}{7} \times 0.25 \times 1.5 \times\left(\frac{2}{3}\right)^2=6.6 \mathrm{~N}\)

If \(\mathrm{T}_{\max }=200 \mathrm{~N}\), then from \(\mathrm{T}_{\max }=\frac{\mathrm{mv}_{\max }^2}{\mathrm{r}}\)

⇒ \(\mathrm{v}_{\max }^2=\frac{\mathrm{T}_{\max } \times \mathrm{r}}{\mathrm{m}}=\frac{200 \times 1.5}{0.25}=1200\)

⇒ \(\mathrm{v}_{\max }=\sqrt{1200}=34.6 \mathrm{~m} / \mathrm{s}\)

Question 26. Explain why

1. A horse cannot pull a cart and run in empty space,
2. Passengers are thrown forward from their seats when a speeding bus stops suddenly,
3. It is easier to pull a lawn mover than to push it,
4. A cricketer moves his hands backward while holding a catch.

Solution:

1. While trying to pull a cart, a horse pushes the ground backward with a certain force at an angle. The ground offers an equal reaction in the opposite direction, on the feel of the horse. The forward component of this reaction is responsible for the motion of the cart. In empty space, there Is no reaction, and hence, a horse cannot pull the cart and run.
2. This is due to the “inertia of motion”. When the speeding bus stops suddenly, the lower part of the body in contact with the seats stops. The upper part of the bodies of the passengers tends to maintain a uniform motion. Hence, the passengers are thrown forward.
3. While pulling a lawn mover, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the mover. While pushing a lawn mover, force is applied downwards along the handle. The vertical component of this force is downwards and increases the effective weight of the mover. As the effective weight is lesser in the case of pulling than in the case of pushing, therefore, “pulling is easier than pushing”.
4. While holding a catch, the impulse received by the hands, F x t = change in linear momentum of the ball is constant. By moving his hands backward, the cricketer increases the time of impact (t) to complete the catch. As t increases, F decreases and as a reaction, his hands are not hurt severely.

Question 27. A stream of water flowing horizontally with a speed of 15 m s^-1 pushes out of a tube of cross-sectional area 10^-2 m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Solution:

Here, v = 15 ms^-1

Area of cross-section, a = 10^-2 m²

The volume of water pushing out/sec = a x v = 10^-2 x 15 m³ s^-1

As the density of water is 10³ kg/m³, therefore, mass of water strikes the wall per sec.

m = (15 x 10^-2) x 10³ = 150 kg/s.

As \(\mathrm{F}=\frac{\text { change in linear momentum }}{\text { time }}\)

∴ \(\mathrm{F}=\frac{\mathrm{m} \times \mathrm{v}}{\mathrm{t}}=\frac{150 \times 15}{1}=2250 \mathrm{~N}\)

KSEEB Physics Class 11 Laws Of Motion Notes

Question 28. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass of m. Give the magnitude and direction of

1. The force on the 7th coin (counted from the bottom) due to all the coins on its top,
2. The force on the 7th coin by the eighth coin,
3. The reaction of the 6th coin on the 7th coin.

Solution:

1. The force on the 7th coin is due to the weight of the three coins lying above it. Therefore, F = (3 m) kgf = (3 mg) N

where g is the acceleration due to gravity. This force acts vertically downwards.

2. The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e. F = 2 m + m = (3 m) kgf = (3 mg) N

The force acts vertically downwards.

3. The sixth coin is under the weight of four coins above it.

Reaction, r = -F = -4 m (kgf) =- (4 mg) N

The Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.

Question 29. An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Solution:

Here θ = 15°

v \(=720 \mathrm{~km} / \mathrm{h}=\frac{720 \times 1000}{60 \times 60}=200 \mathrm{~ms}^{-1}\);

g = \(9.8 \mathrm{~ms}^{-2} \text {; from } \tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}} \)

⇒ \(\mathrm{v}^2=\mathrm{rg} \tan \theta\)

∴ r \(=\frac{\mathrm{v}^2}{\mathrm{~g} \tan \theta}=\frac{(200)^2}{9.8 \times \tan 15^{\circ}}\)

= \(15232. \mathrm{m}=15.232 \mathrm{k} \cdot \mathrm{m} \text {. }\)

Question 30. A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h The mass of the train is 10⁶ kg. What provides the centripetal force required for, this purpose-The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Solution:

The centripetal force is provided by the lateral thrust exerted by the rails on the wheels. By Newton’s 3rd law, the train exerts an equal and opposite thrust on the rails causing its wear and tear.

Obviously, the outer rail will wear out faster due to the larger force exerted by the train on it.

Here, \(v=54 \mathrm{~km} / \mathrm{h}=\frac{54 \times 1000}{60 \times 60}=15 \mathrm{~m} / \mathrm{s}\);

g = \(9.8 \mathrm{~ms}^{-2}\)

As \(\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{15 \times 15}{30 \times 9.8}=0.76 ;\)

∴ \(\theta=\tan ^{-1} 0.76=37.4^{\circ}\)

Question 31. A block of mass 25 kg Is raised by a 50 kg man in two different ways as shown. What Is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

Solution:

Here, the mass of the block, m = 25 kg

Mass of man, M = 50 kg

Force applied to lift the block

F = mg = 25 x 9.8 = 245 N

Weight of man W = Mg = 50 x 9.8 = 490 N

1. When a block is raised by the man as shown, force is applied by the man in the upward direction. This increases the apparent weight of the man. Hence action on the floor.

W’ = W + F = 490 + 245 = 735 N

When a block is raised by a man as shown, force is applied by the man in the downward direction. This decreases the apparent weight of the man. Hence, action on the floor in this case would be W’ = W – F = 490 – 245 = 245 N.

As the floor yields to a normal force of 700 N, mode (2) has to be adopted by the man to lift the block.

Question 32. A monkey of mass 40 kg climbs on a rope that can stand a maximum tension of 600 N. In which of the following cases will the rope break the monkey

1. Climbs up with an acceleration of 6 ms^-2
2. Climbs down with an acceleration of 4 ms^-2
3. Climbs up with a uniform speed of 5 ms^-1
4. Fell down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Solution:

Here, the mass of the monkey, m = 40 kg.

Maximum tension the rope can stand, T = 600 N.

In each case, the actual tension in the rope will be equal to the apparent weight of the monkey (R),

The rope will break when R exceeds T.

1. When monkey climbs up with a = 6 ms^-2, R = m (g + a) = 40 (10 + 6) = 640 N (which is greater than T).

Hence the rope will break.

2. When the monkey climbs down with a = 4 ms^-2

R = m (g – a) = 40 (10 – 4) = 240 N, which is less than T

∴ The rope will not break.

3. When the monkey climbs up with a uniform speed v = 5 ms^-1,

its acceleration a = 0

R = mg = 40 x 10 = 400 N, which is less than T

∴ The rope will not break.

4. When the monkey falls down the rope nearly freely under gravity, a = g

R = m (g – a) = m (g – g) = 0 (Zero.)

Hence the rope will not break.

Question 33. A 70 kg man stands In contact against the Inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical avis with 200 rev/mln. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Solution:

Here, m = 70 kg, r = 3 m

n = 200, rpm = \(\frac{200}{60}\) rps, μ = 0.15, ω= ?

The horizontal force N by the wall on the man provides the necessary centripetal force = m r ω². The frictional force (f) in this case is vertically upwards opposing the weight (mg) of the man.

After the floor is removed, the man will remain stuck to the wall, when mg = f < μ N, i.e. mg < μ m r ω² or g < μ r ω²

∴ The minimum angular speed of rotation of the cylinder is \(\omega=\sqrt{\frac{g}{\mu r}}=\sqrt{\frac{10}{0.15 \times 3}}\)

= \(4.7 \mathrm{rad} / \mathrm{s}\)

Question 34. A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for \(\omega \leq \sqrt{g / R},\). What is the angle made by the radius vector joining the center to the bead with the vertically downward direction form = \(\omega \leq \sqrt{2g / R},\) Neglect friction?
Solution:

In Figure, we have shown that the radius vector joining the bead to the center of the wire makes an angle 0 with the verticle downward direction. If N is a normal reaction, then as is clear from the figure,

⇒ \(\mathrm{mg}=\mathrm{N} \cos \theta\)….(1)

⇒ \(\mathrm{mr} \omega^2=\mathrm{N} \sin \theta\)…..(2)

or \(\mathrm{m}(\mathrm{R} \sin \theta) \omega^2=\mathrm{N} \sin \theta \text { or } \mathrm{m} \mathrm{R}^2=\mathrm{N}\)

from (1), \(\mathrm{mg}=\mathrm{m} R \omega^2 \cos \theta\)

or \(\cos \theta=\frac{\mathrm{g}}{\mathrm{R \omega}^2}\) …..(3)

As |cos θ| ≤ 1, therefore, the bead will remain at its

lowermost point for \(\frac{g}{R \omega^2} \leq 1 \text {, or } \omega \leq \sqrt{\frac{g}{R}}\)

When \(\omega=\sqrt{\frac{2 g}{R}}\) from (3), \(\cos\theta=\frac{g}{R}\left(\frac{R}{2 g}\right)=\frac{1}{2}\)

∴ \(\theta=60^{\circ}\)

KSEEB Class 11 Physics Solutions For Chapter 4 Motion in a Plane Very Short Answer Questions

Question 1. Write the equation for the horizontal – range covered by a projectile and specify
when it will be maximum.
Answer:

Range of a projectile (R) = \(\frac{u^2 \sin 2 \theta}{g}\)

When θ = 45° Range is maximum. (sin 90° = 1)

Maximum Range (R_max) = \(\frac{u^2}{g}\)

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. The vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with the X-axis?
Answer:

Let R be a vector.

Vertical component = R sin θ;

Horizontal component = R cos θ

∴ R sin θ = R cos θ

So sin θ = cos θ ⇒ θ = 45°

Question 3. A vector V makes an angle θ with the horizontal. The vector is rotated through an angle α. Does this rotation change the vector V?
Answer:

Magnitude of vector = V ;

Let the initial angle with horizontal = θ

Angle rotated = α

So new angle with horizontal = θ + α

Now horizontal component, V_α = V cos (θ + α)

Vertical component, V_y = V sin (θ + α)

Magnitude of vector, V = \(\sqrt{V_x^2+V_y^2}=V\)

So rotating the vector does not change its magnitude.

KSEEB Class 11 Physics Motion In A Plane

Question 4. Two forces of magnitude 3 units and 5 units act at 60 with each other, What is the magnitude of their resultant?
Answer:

Given \(\bar{P}\) = 3 units, \(\bar{Q}\) = 5 units and θ = 60°

∴ R = \(\sqrt{P^2+Q^2+2 P Q \cos \theta}\)

= \(\sqrt{3^2+5^2+2(3)(5) \cos 60^{\circ}}\)

= \(\sqrt{9+25+30 \times \frac{1}{2}}=\sqrt{49}=7 \text { units. }\)

Question 5. A = \(\vec{i}\) + \(\vec{j}\). What is the angle between the vector and the X-axis?
Answer:

Given that, \(\vec{A}\) = \(\vec{i}\) + \(\vec{j}\)

|\(\vec{A}\)| = \(\sqrt{(1)^2+(1)^2}=\sqrt{2}\)

If ‘θ’ is the angle made by the vector with X-axis then, \(\cos \theta=\frac{A_x}{|\vec{A}|} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}\)

Question 6. When two right-angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant?
Answer:

Given \(\vec{P}\) = 7 units; \(\vec{Q}\) = 24 units;  θ = 90°

∴ \(\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}\)

∴ \(\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}=\sqrt{7^2+24^2}=25\) units.

KSEEB Physics Class 11 Motion In A Plane Notes

Question 7. If \(\vec{P}\) = 2i + 4j + 14k and \(\vec{Q}\) = 4i + 4j + 10k, find the magnitude of \(\vec{P}\) + \(\vec{Q}\).
Answer:

Given \(\overline{\mathrm{P}}=2 \overline{\mathrm{i}}+4 \overline{\mathrm{j}}+14 \overline{\mathrm{k}}\) and

⇒ \(\bar{Q}=4 \overline{\mathrm{i}}+4 \overline{\mathrm{j}}+10 \overline{\mathrm{k}}\)

⇒ \(\overline{\mathrm{P}}+\overline{\mathrm{Q}}=6 \overline{\mathrm{i}}+8 \overline{\mathrm{j}}+24 \overline{\mathrm{k}}\)

Magnitude of \(|\overline{\mathrm{P}}+\overline{\mathrm{Q}}|=\sqrt{6^2+8^2+24^2}\)

= \(\sqrt{36+64+576}=\sqrt{676}=26\) units.

Question 8. Can a vector of magnitude zero have non-zero components?
Answer:

A vector with zero magnitude cannot have non-zero components, because the magnitude of a given vector \(\bar{V}=\sqrt{V_x^2+V_y^2}\) must be zero. This is possible only when \(V_x^2\) and \(V_y^2\) are zero.

Question 9. What is the acceleration of a projectile at the top of its trajectory?
Answer:

At Highest point acceleration, a = g. In projectile motion acceleration will always act towards the centre of the earth. It is irrespective of its position, whether it is at the highest point or somewhere.

Question 10. Can two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors add up to give the zero vector?
Answer:

No. Two unequal vectors can never give zero vector by addition. But three unequal vectors when added may give zero vector.

Motion In A Plane Solutions KSEEB Class 11 Physics Chapter 4 Short Answer Questions

Question 1. State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.
Answer:

Parallelogram Law: If two vectors are represented by the two adjacent sides of a parallelogram, then the diagonal passing through the intersection of given vectors represents their resultant both in direction and magnitude.

Proof: Let \(\overline{\mathrm{P}}\) and \(\overline{\mathrm{Q}}\) be two adjacent vectors ‘θ’ be the angle between them. Construct a parallelogram OACB as shown in the figure.

Extend the line OA and draw a normal D from C. The diagonal OC = the resultant \(\overline{\mathrm{R}}\) both in direction and magnitude

In figure OCD = right angle triangle ⇒ OC² = OD² + DC²

But OD = OA + AD = \(\overline{\mathrm{P}}\) + \(\overline{\mathrm{Q}}\) cos θ and CD = Q sin θ and \(\overrightarrow{\mathrm{OC}}=R\)

∴ \(R^2=[P+Q \cos \theta]^2+[Q \sin \theta]^2\)

⇒ \(\overline{\mathrm{R}}^2=\overline{\mathrm{P}}^2+\overline{\mathrm{Q}}^2 \cos ^2 \theta+2 \overline{\mathrm{PQ}} \cos \theta+\overline{\mathrm{Q}}^2 \sin ^2 \theta\)

∴ \(\overline{\mathrm{R}}^2=\overline{\mathrm{P}}^2+\overline{\mathrm{Q}}^2\left(\sin ^2 \theta+\cos ^2 \theta\right)+2 \overline{\mathrm{PQ}} \cos \theta\)

But \(\sin ^2 \theta+\cos ^2 \theta=1\)

∴ \(\mathrm{R}^2=[\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta\).

∴ Resultant vector, \(\overline{\mathrm{R}}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ}}\)

The angle of resultant with adjacent side ‘α’

In triangle OCD, \(\tan \alpha=\frac{C D}{O D}\) (OD = OA + AD)

⇒ \(\text{Tan} \alpha=\frac{C D}{O A+A D}=\frac{Q \sin \theta}{P+Q \cos \theta}\)

(because A D = \(Q \cos \theta\))

Angle of \(\overline{\mathrm{R}}\) with adjacent side, \(\alpha=\tan ^{-1}\left[\frac{Q \sin \theta}{P+Q \cos \theta}\right]\)

Question 2. What is relative motion? Explain it.
Answer:

Relative velocity is the velocity of a body with respect to another moving body.

Relative velocity in two-dimensional motion

Let two bodies A and B are moving with velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) then relative velocity of A with respect to B is \(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}=\overrightarrow{\mathrm{V}}_{\mathrm{A}}-\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)

The relative velocity of B with respect to A is \(\overrightarrow{\mathrm{V}}_{\mathrm{BA}}=\overrightarrow{\mathrm{V}}_B-\overrightarrow{\mathrm{V}}_{\mathrm{A}}\)

Procedure To Find Resultant: To find relative velocity in two-dimensional motion use vectorial subtraction of V_A or V_B. Generally to find relative velocity one vector \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) is reversed (as the case may be) and parallelogram is constructed. Now resultant of that parallelogram is equal to \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) (one vector is reversed VA is taken as –\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) is taken as –\(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\))

In the figure relative velocity of B with respect to A is V_BA = V_R = V_B -V_A

Question 3. Show that a boat must move at an angle with respect to river water in order to cross the river in minimum time.
Answer:

The motion of a boat in a river: Let a boat travel with a speed of V_bE in still water with respect to earth. It is used to cross a river which flows with a speed of V_WE with respect to earth. Let the width of the river be W.

We can cross the river in two different ways.

1. In shortest path
2. In the shortest time.

To cross the river in the shortest time:

Shortest time, t = \(\mathrm{t}=\frac{\text { width of river }}{\text { velocity of boat }}=\frac{W}{V_{\mathrm{bE}}}\)

To cross the river in the shortest time boat must be rowed along the width of the river i.e., the boat must be rowed perpendicular to the bank or 90° with the flow of water.

Because the width of the river is the shortest distance. So velocity must be taken in that direction to obtain the shortest time.

In this case, V_bE and V_WE are perpendicular and the boat will travel along AC. The distance BC is called drift. So to cross the river in the shortest time angle with the flow of water = 90°.

KSEEB Class 11 Physics Chapter 4 Solved Examples

Question 4. Define unit vector, null vector and position vector.
Answer:

Unit Vector: A vector whose magnitude is one unit is called a unit vector.

Let \(\overline{\mathrm{a}}\) is a given vector then the unit vector of \(\bar{a}=\frac{\bar{a}}{|\bar{a}|}=\hat{a}\)

When \(\bar{a}\) ≠ 0 or \(\frac{\bar{a}}{|\bar{a}|}\) = unit vector of \(\bar{a}\). It is denoted by a \(\hat{a}\)

Null Vector: A vector whose magnitude is zero is called a null vector. But it has direction.

For a null vector, the origin and terminal point are the same.

Example: Let \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}=\overline{0}\). Here magnitude of \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}=\overline{0}\). But still, it has direction perpendicular to the plane of \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{A}}\).

Position Vector: Any vector in space can be represented by the linear combination of \(\overline{\mathrm{i}}, \overline{\mathrm{j}}\) and \(\overline{\mathrm{k}} \text {. }\).

Let ‘O’ is the origin then \(\overline{\mathrm{OP}}\) is represented as \(\overline{\mathrm{OP}}=x \bar{i}+y \bar{j}+z \bar{k}\) where x, y and z are magnitudes of \(\overline{\mathrm{OP}}\) along \(\bar{i}, \bar{j}\) and \(\overline{\mathrm{k}}\) axis.

Magnitude of \(\overrightarrow{\mathrm{OP}}=\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\)

Question 5. If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), prove that the angle between \(\vec{a}\) and \(\vec{b}\) is 90°.
Answer:

Let \(\vec{a}\), \(\vec{b}\) are the two vectors.

Sum of vectors = \(\bar{a}+\bar{b}=\sqrt{a^2+b^2+2 a b \cos \theta}\)

Difference of vectors = \(\bar{a}-\bar{b}=\sqrt{a^2+b^2-2 a b \cos \theta}\)

Given \(|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|\)

⇒ \(\sqrt{a^2+b^2+2 a b \cos \theta}\)

= \(\sqrt{a^2+b^2-2 a b \cos \theta}\) by squaring on both sides, \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ab} \cos \theta=\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab} \cos \theta\)

∴ \(4 \mathrm{ab} \cos \theta=0\) or \(\theta=90^{\circ}\)

So if \(|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|\) then angle between \(\bar{a}\) and \(\bar{b}\) is \(90^{\circ}\).

Motion In A Plane Solutions KSEEB Class 11 Physics

Question 6. Show that the trajectory of an object thrown at a certain angle with the horizontal is a parabola.
Answer:

Projectile: A body thrown into the air same angle as the horizontal, (other than 90°) its motion under the influence of gravity is called a projectile. The path followed by it is called a trajectory.

Let a body be projected from point O, with velocity ‘u’ at an angle θ with horizontal. The velocity ‘u’ can be resolved into two rectangular components u_x and u_y along X-axis and Y-axis.

u_x = u cos θ and u_y = u sin θ

After time t, Horizontal distance travelled x = u cos θ t……(1)

After a time sec; vertical displacement y = u sinθ • t – \(\frac{1}{2}\) gt²

y = \(u \sin \theta \cdot t-\frac{1}{2} g t^2\)

From (1) t = \(\frac{x}{u \cos \theta}\)

y = \(\frac{u \sin \theta \cdot x}{u \cos \theta}-\frac{1}{2} g \cdot \frac{x^2}{u^2 \cos ^2 \theta}\)

⇒ \(y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}\)

Let \(\tan \theta=\mathrm{A}\) and \(\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{B}\) then y = \(A x-B x^2\)

Let \(\tan \theta=\mathrm{A}\) and \(\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{B}\) then \(y=A x-B x^2\)

The above equation represents a “parabola”. Hence the path of a projectile is a parabola.

Question 7. Explain the terms average velocity and instantaneous velocity. When are they equal?
Answer:

Average Velocity: It is the ratio of total displacement to total time taken.

Average velocity = \(\frac{\text { total displacement }}{\text { total time taken }}\)

= \(\frac{x_2-x_1}{t_2-t_1}\)

Average velocity is independent of the path followed by the particle. It just deals with the initial and final positions of the body.

Instantaneous Velocity: The velocity of a body at any particular instant of time is defined as instantaneous velocity.

Mathematically \(\frac{\mathrm{dx}}{\mathrm{dt}}=\underset{\Delta t \rightarrow 0}{\text{Lt}} \frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\) is defined as instantaneous velocity.

For a body moving with uniform velocity its average velocity = Instantaneous velocity.

Question 8. Show that the maximum height and range of a projectile are \(\frac{\mathrm{U}^2 \sin ^2 \theta}{2 \mathrm{~g}}\) and \(\frac{\mathrm{U}^2 \sin 2 \theta}{\mathrm{g}}\) respectively where the terms have their regular meanings.
Answer:

Let a body be projected with an initial velocity ’u’ and with an angle θ to the horizontal.

Initial velocity along x direction, u_x = u cos θ

Initial velocity along y direction, u_y = u sin θ

Maximum Height: The vertical distance covered by the projectile until its vertical component becomes zero.

from equation v² – u² = 2as ……..(1)

at maximum height v = 0

u = u_y = u sin θ

a = -g

S = H

From equation (1)

0² – u² sin²θ = 2 (- g) H

– u² sin²θ = – 2 gH

∴ H = \(\frac{u^2 \sin ^2 \theta}{2 g}\)

KSEEB Class 11 Physics Chapter 4 Motion In A Plane

Horizontal Range: It is the distance covered by the projectile along the horizontal between the point of projection to the point on the ground, where the projectile returns again.

It is denoted by R. The horizontal distance covered by the projectile at the time of flight is called horizontal range.

Therefore, R = u cos θ x t.

R = \(\mathrm{u} \cos \theta \times \frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{\mathrm{u}^2(2 \sin \theta \cos \theta)}{\mathrm{g}}\),

But \(2 \sin \theta \cos \theta=\sin 2 \theta\),

∴ Range(R) = \(\frac{u^2 \sin 2 \theta}{g}\)

Angle Of Projection For Maximum Range: For a given velocity of projection, the horizontal range will be maximum, when sin 2θ = 1.

∴ Angle of projection for maximum range is 2θ = 90° or θ = 45°

∴ \(R_{\max }=\frac{u^2}{g}\)

Question 9. If the trajectory of a body IN parabolic In one reference frame, can It be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else can it be?
Answer:

Yes. According to Newton’s first law, a body at rest or a body moving with uniform velocity is treated as the same. Both of them belong to an inertial frame of reference.

If a frame (say 1) is moving with uniform velocity with respect to others, then that second frame must be at rest or it maintains a constant velocity concerning the first.

So both frames are inertial frames. So if the trajectory of a body in one frame is a parabola, then the trajectory of that body in another frame is also a parabola.

Question 10. A force 2i + j – k Newton acts on a body which is initially at rest. At the end of 20 seconds, the velocity of the body is 4i + 2j – 2k ms^-1. What is the mass of the body?
Answer:

Force, F = 2i + j – k

time, t = 20

Initial velocity, u = 0

Final velocity, v = 4i + 2j – 2k = 2(2i + j – k)

Acceleration, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\)

= \(\frac{2(2 i+j-k)-0}{20}=\frac{2 i+j-k}{10}\)

Mass of the body, m = \(\frac{F}{a}\)

= \(\frac{2 i+j-k}{(2 i+j-k) / 10}=10 \mathrm{~kg}\)

Chapter 4 Motion In-Plane Problems In KSEEB Physics Class 11

Question 1. Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30 km/h, while ship B is heading in a direction 60° west of north at a speed of 20 km/h.

1. Determine the magnitude of the velocity of ship B relative to ship A.
2. What will be their distance of closest approach?

Answer:

The velocity of A 30 kmph due North

∴ V_A=30 \(\hat{j}\)

Velocity of B = 20 kmph 60° west of North

∴ \(V_B=-20 \sin 60^{\circ}+20 \cos 60^{\circ}=10 \sqrt{3} \hat{\mathrm{i}}+10 \hat{\mathrm{j}}\)

Velocity of B with respect to A = V_B – V_A

= \(-10 \sqrt{3} \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-30 \hat{\mathrm{j}}=-10 \sqrt{3} \hat{\mathrm{i}}-20 \hat{\mathrm{j}}\)

∴ \(\mathrm{V}_R=\left|\mathrm{V}_B-\mathrm{V}_A\right|=\sqrt{100 \times 3+400}\)

= \(10 \sqrt{7} \mathrm{kmph}\)

Shortest Distance: In Δ le ANB shortest distance, AN = AB sin θ

But distance, AB = 10 km

∴ AN = \(10 \times \frac{20}{10 \sqrt{7}}=\frac{20}{\sqrt{7}}=7.56 \mathrm{~km}\)

Motion In A Plane KSEEB Physics

Question 2. If θ is the angle of projection, R is the range, h is the maximum height, and T is the time of flight, then show that

1. tan θ = 4h/R and
2. h = gT²/8

Answer:

1. Given angle of projection = η,

Range, \(R=\frac{u^2 \sin 2 \theta}{g}\)

h = \(\mathrm{h}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \text {. }\)

Time of flight \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\)

= \(\frac{\sin ^2 \theta}{2 \sin 2 \theta}=\frac{\sin \theta \sin \theta}{2 \times 2 \sin \theta \cos \theta}\)

∴ \(\frac{h}{R}=\frac{\tan \theta}{4} \Rightarrow \tan \theta=\frac{4 h}{R}\)

2. \(\mathrm{h}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\)

But \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}} \Rightarrow \frac{\mathrm{u} \sin \theta}{\mathrm{g}}=\mathrm{T}/2\).

multiply with (g/g)

h = \(\frac{u^2 \sin ^2 \theta}{2 g}=\frac{u^2 \sin ^2 \theta}{2 \cdot g^2} \cdot g\)

= \(\left(\frac{T}{2}\right)^2 \times \frac{g}{2}=\frac{g T^2}{8}\)

∴ \(\mathrm{h}=\frac{\mathrm{gT}^2}{8}\) is proved.

Question 3. A projectile is fired at an angle of 60° to the horizontal with an initial velocity of 800 m/s:

1. Find the time of flight of the projectile before it hits the ground.
2. Find the distance it travels before it hits the ground (range).
3. Find the time of flight for the projectile to reach its maximum height.

Answer:

Angle of projection, θ = 60°,

Initial velocity, u = 800 m/s

1. Time of flight, \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\)

= \(\frac{2 \times 800}{9.8} \times \frac{\sqrt{3}}{2}=\frac{800 \times 1.732}{9.8}=141.4 \mathrm{sec}\)

2. Range, \(R =\frac{u^2 \sin 2 \theta}{g}\)

= \(\frac{800 \times 800 \times \sin \left(2 \times 60^{\circ}\right)^{\prime}}{9.8}\)

R = \(\frac{800 \times 800}{9.8} \frac{\sqrt{3}}{2}\)

= \(\frac{800 \times 800}{9.8} \times 0.8660=56.56 \mathrm{~km}\)

3. Time of flight to reach maximum height = \(\frac{T}{2}\)

= \(\frac{800 \times 0.8660}{9.80}=70.7 \mathrm{sec}\)

Motion In A Plane Problems In KSEEB Physics

Question 4. For a particle projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to be √2 times the maximum height reached by it. Show that the angle of projection is tan^-1 √2.
Answer:

The position vector of h (max point) from 0, is \(\frac{\mathrm{R}}{2}=\frac{1}{2} \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\)

But R/2 = √2 h_max(given)

∴ \(\frac{1}{2} \frac{u^2 \sin 2 \theta}{g}=\frac{\sqrt{2} \mathrm{u}^2 \sin ^2 \theta}{2 g} \)

∴ \(2 \sin \theta \cos \theta=\sqrt{2} \sin \theta \cdot \sin \dot{ }\)

⇒ \(\frac{2}{\sqrt{2}}=\frac{\sin \theta}{\cos \theta} \Rightarrow \tan \theta=\sqrt{2}\)

∴ Angle of projection, \(\theta=\tan ^{-1} \sqrt{2}\)

Question 5. An object is launched from a cliff 20 m above the ground at an angle of 30° above the horizontal with an initial speed of 30 m/s. How far horizontally does the object travel before landing on the ground? (g = 10 m/s²).
Answer:

Height of cliff = 20m

The angle of projection, θ = 30°

Velocity of projection, u = 30 m/s

Total horizontal distance travelled = OC = OB’+ B’C

But OB’ = AB = Range R = \(\frac{u^2 \sin 2 \theta}{\mathrm{g}}\)

∴ R = \(\frac{30 \times 30 \sin 60^{\circ}}{10}=90 \frac{\sqrt{3}}{2}=45 \sqrt{3} \rightarrow(1)\)

2. Distance B’C = Range of a horizontal projectile.

∴ Range = u cos θ t

u \(\cos \theta=30 \cdot \frac{\sqrt{3}}{2}=15 \sqrt{3}\)

Time taken to reach the ground, t =?

Given \(S_y=20, \quad u_y=u \sin \theta=30 \sin 30^{\circ}\)

= \(15 \mathrm{~m} / \mathrm{s}\)

from \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} a t^2\)

∴ \(\mathrm{s}_{\mathrm{y}}=20=15 \mathrm{t}+\frac{10}{2} \mathrm{t}^2 \Rightarrow 5 \mathrm{t}^2+15 \mathrm{t}-20=0\)

or \(t^2+3 t-4=0\) or \((t+4)(t-1)=0\)

∴ \(\mathrm{t}=-4\) or \(\mathrm{t}=1\)

∴ \(\mathrm{t}\) is Not -ve use t=1

∴ Range = \(\mathrm{u} \cdot \cos \theta \cdot \mathrm{t}=15 \sqrt{3} \times 1 \rightarrow(2)\)

Total distance travelled before reaching the equation (1) + (2) ground = 45√3 + 15√3 =60√3 m.

KSEEB Class 11 Physics Motion In A Plane Questions And Answers

Question 6. ‘O’ is a point on the ground chosen as origin. A body first suffers a displacement of 10√2 mm North-East, next 10 m North and finally 10√2 North-West. How far it is from its origin?
Answer:

1. 10√2 m North-East
2. 10 m North
3. 10√2 m North West

From the figure total displacement from the origin, ‘O’ is OC

But OC = OA’ + A’B’ + B’C = 10 + 10 + 10 = 30 m.

Question 7. From a point on the ground, a particle is projected with initial velocity u, such that its horizontal range is maximum. Find the magnitude of average velocity during its ascent.
Answer:

Velocity of projection = u.

Range is maximum ⇒ θ = 45°

During time of ascent ⇒ when h = h_max ⇒ u_x = v_x= u · cos θ

Average velocity, \(V_A=\sqrt{V_x^2+V_y^2}\)

V_x = Average velocity along X-axis = \(\frac{\mathrm{u} \cdot \cos \theta+\mathrm{u} \cos \theta}{2}\)

⇒ \(\mathrm{v}_{\mathrm{x}}=\mathrm{u} \cdot \cos \theta=\mathrm{u} \cdot \cos 45^{\circ}=\mathrm{u} / \sqrt{2} \longrightarrow(1)\)

Average velocity along \(\mathrm{Y} \text {-axis }=\frac{\mathrm{u}_{\mathrm{y}}+\mathrm{v}_{\mathrm{y}}}{2}\)

⇒ \(\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta=\mathrm{u} / \sqrt{2}, \mathrm{v}_{\mathrm{y}}=0\left(\text { at }_{\mathrm{h}_{\max }}\right)\)

∴ \(\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{u} / \sqrt{2}}{2}=\frac{\mathrm{u}}{2 \sqrt{2}} \longrightarrow(2)\)

Average velocity during the time of ascent

= \(\sqrt{\left(\frac{u}{\sqrt{2}}\right)^2+\left(\frac{u}{2 \sqrt{2}}\right)^2}\)

Average velocity

= \(\sqrt{\frac{u^2}{2}+\frac{u^2}{4 \times 2}}=\sqrt{\frac{4 u^2+u^2}{8}}=\frac{u \sqrt{5}}{2 \sqrt{2}}\)

Question 8. A particle is projected from the ground with some initial velocity making an angle of 45° with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10 m from the point of projection. Find the initial speed of projection (g = 10 m/s²).
Answer:

Angle of projection = 45°

Vertical height, h_y = 7.5 m

Horizontal distance, h_x = 10 m

From figure \(\tan \theta=\frac{y}{x}=\frac{7.5}{10}=3 / 4, \cos \theta=\frac{4}{5}\)

⇒ \(\mathrm{s}_{\mathrm{x}}=\mathrm{u} \cdot \cos \theta \cdot \mathrm{t} \Rightarrow 10=\mathrm{u} \cdot \frac{1}{\sqrt{2}} \cdot \mathrm{t}\)

t = \(\frac{10 \sqrt{2}}{\mathrm{u}}\)…..(1)

⇒ \(\mathrm{s}_{\mathrm{y}}=7.5=(\mathrm{u} \sin \theta) \mathrm{t}-\frac{1}{2} g \mathrm{t}^2\)

7.5 = \(\frac{\mathrm{u}}{\sqrt{2}} \frac{10 \sqrt{2}}{\mathrm{u}}-\frac{1}{2} \times \frac{10 \times 10 \times 10 \times 2}{\mathrm{u}^2}\)

7.5 = \(10-\frac{1000}{\mathrm{u}^2} \Rightarrow-2.5=\frac{-1000}{\mathrm{u}^2}\)

∴ \(\mathrm{u}^2=\frac{1000}{2.5}=400\)

∴ \(u^2=400 \Rightarrow u=20 \mathrm{~m} / \mathrm{s}\)

Question 9. The wind is blowing from the south at 5 ms^-1. To a cyclist, It appears to be blowing from the cast at 5 ms^-1. Show that the velocity of the cyclist is ms^-1 towards the north-cast.
Answer:

The direction of wind is South to North 5 m/s.

The apparent direction is from East to West at 5 m/s.

This is relative velocity.

To find the velocity of the cyclist reverse the direction of the resultant vector OB and find the resultant

∴ Velocity of cyclist = \(\sqrt{5^2+5^2+0}\)

= \(5 \sqrt{2} \mathrm{~m}\)

Motion In A Plane KSEEB Physics

Question 10. A person walking at 4 m/s finds raindrops falling slantwise into his face at a speed of 4 m/s at an angle of 30° with the vertical. Show that the actual speech of the raindrops is 4 m/s.
Answer:

Velocity of man = 4 m/sec

The apparent velocity of raindrop = 4 m/sec with θ = 30° with vertical.

This is relative velocity V_B.

1. Velocity of man = 4 \(\hat{i}\)

Velocity of rain = 4 m/s, 30° with vertical

∴ It is represented by = -4 sin 30 i – 4.cos 30° \(\hat{j}\)

= \(-4 \cdot \frac{1}{2} \hat{\mathrm{i}}+4 \frac{\sqrt{3}}{2} \text { say } V_R=-2 \hat{\mathrm{i}}+2 \sqrt{3} \hat{\mathrm{j}}\)

But V_R = V_B – V_A where V_B is the actual velocity of rain

⇒ \(V_B-V_A+V_A=V_B\)

= \(-2 \hat{i}+2 \sqrt{3} \hat{j}+4 \hat{\mathrm{i}}=2 \hat{\mathrm{i}}+2 \sqrt{3} \hat{\mathrm{j}}\)

∴ \(\left|V_B\right|=\sqrt{4+4 \times 3}=\sqrt{16}=4 \mathrm{~m} / \mathrm{s}\)

Question 11. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms_^-1 can go without hitting the ceiling of the hall?
Solution:

Here, u = 40 ms^-1; H = 25m, R = ?

Let θ be the angle of projection with the horizontal direction to have the maximum range, with maximum height = 25 m.

Maximum height, \(\mathrm{H} \Rightarrow 25=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\)

= \(\frac{(40)^2 \sin ^2 \theta}{2 \times 9.8}\)

or \(\sin \theta=\left(\frac{25 \times 2 \times 9.8}{40^2}\right)=0.5534\)

= \(\sin 33.6^{\circ}\) or \(\theta=33.6^{\circ}\)

Horizontal range, \(\mathrm{R}=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\)

= \(\frac{(40)^2 \times \sin \left(2 \times 33.6^{\circ}\right)}{9.8}\)

= \(\frac{1600 \times 0.9219}{9.8}=150.5 \mathrm{~m}\)

Question 12. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds, what is the magnitude and direction of acceleration of the stone?
Solution:

Here, r = 80 cm = 0.8 m; v = 14/25 s^-1.

∴ \(\omega=2 \pi v=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} \text { rad. s } \mathrm{s}^{-1} \text {. }\)

The centripetal acceleration, \(\mathrm{a}=\omega^2 \mathrm{r}=\left(\frac{88}{25}\right)^2 \times 0.80=9.90 \mathrm{~m} / \mathrm{s}^2\)

The direction of centripetal acceleration is along the string directed towards the centre of the circular path.

KSEEB Class 11 Physics Motion In A Plane Key Concepts

Question 13. An aircraft executes a horizontal loop of radius 1 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution:

Here, r = 1 km = 1000 m;

v = 900 km h^-1 = 900 x (1000m) x (60 x 60s)^-1 = 250 ms^-1

Centripetal acceleration, \(\mathrm{a}=\frac{\mathrm{v}^2}{\mathrm{r}}=\frac{(250)^2}{1000}\)

Now, \(\frac{\mathrm{a}}{\mathrm{g}}=\frac{(250)^2}{1000} \times \frac{1}{9.8}=6.38\)

Question 14. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Solution:

Figure O is the observation point at the ground. A and B are the positions of aircraft for which ∠AOB = 30°.

Draw a perpendicular OC on AB. Here OC = 3400 m and ∠AOC = ∠COB = 15°. Time taken by aircraft from A to B is 10 s.

In ΔAOC, AC = OC tan 15° = 3400 x 0.2679

AC = 910.86 m.

AB = AC + CB = AC + AC = 2 AC

= 2 x 910.86 m

Speed of the aircraft, v = \(\frac{\text { distance } A B}{\text { time }}\)

= \(\frac{2 \times 910.86}{10}=182.17 \mathrm{~ms}^{-1}=182.2 \mathrm{~ms}^{-1}\).

Question 15. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed and neglect air resistance.
Solution:

Horizontal range, R= \(\frac{u^2 \sin 2 \theta}{g}\)

or, 3 = \(\frac{u^2 \sin 60^{\circ}}{g}\) or \(\frac{u^2}{g}=2 \sqrt{3}\)

3 = \(\frac{u^2}{g} \times \frac{\sqrt{3}}{2}\)

2\(\sqrt{3}=\frac{u^2}{g}\)

∴ \(R_{\max }=2 \sqrt{3}\)

∴ \(R_{\max }=3.464 \mathrm{~m} \).

KSEEB Class 11 Physics Solutions For Motion in a Straight Line Chapter 3 Motion In A Straight Line Very Short Answer Questions

Question 1. The states of motion and rest are relative Explain.
Answer:

Rest: If the position of a body does not change with respect to its surroundings, it is said to be at “rest”.

Motion: If the position of a body changes with respect to its surroundings, it is said to be in “motion”.

By definitions rest and motion are relative with respect to surroundings.

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. How is average velocity different from instantaneous velocity?
Answer:

Average Velocity: It is the ratio of total displacement to total time taken. It is independent of the path of the body.

∴ Average velocity = \(\frac{\mathrm{s}_2-\mathrm{s}_1}{\mathrm{t}_2-\mathrm{t}_1}(\mathrm{Or}) \mathrm{V}_{\text {avg }}=\frac{\mathrm{ds}}{\mathrm{dt}}\)

The velocity of a particle at a particular instant of time is known as instantaneous velocity. Here time interval is very small.

V = \(\frac{dx}{dt}\)

Only in uniform motion, instantaneous velocity = average velocity. For all other, cases instantaneous velocity may differ from average velocity.

Question 3. Give an example where the velocity of an object is zero but its acceleration is not zero.
Answer:

In the case of a Vertically Projected Body at maximum height its velocity v = 0. But acceleration due to gravity ’g’ is not zero

So even though velocity v = 0 ⇒ acceleration is not zero.

Question 4. A vehicle travels half the distance L with speed v₁ and the other half with speed v₂ What is the average speed?
Answer:

The average speed of a vehicle for the two  equal parts \(\frac{L}{2}\), \(\frac{L}{2}\) is

⇒ \(V_{\text {avg }}=\frac{\text { Total distance }}{\text { Total time }}=\frac{\frac{L}{2}+\frac{L}{2}}{t_1+t_2}\) (because \(t=\frac{L / 2}{V}\))

⇒ \(V_{\text {avg }}=\frac{L}{\left(\frac{\frac{L}{2}}{V_1}+\frac{\frac{L}{2}}{V_2}\right)} \)

= \(\frac{L}{\frac{L}{2}\left(\frac{1}{V_1}+\frac{1}{V_2}\right)}\)

= \(\frac{1}{\frac{1}{2}\left(\frac{V_1+V_2}{V_1 V_2}\right)}\)

∴ \(V_{\text {avg }}=\frac{2 V_1 V_2}{V_1+V_2}\)

Motion In A Straight Line Solutions KSEEB Class 11 Physics

Question 5. A lift coming down is just about to reach the ground floor. Taking the ground floor as the origin and positive direction upwards for all quantities, which one of the following is correct?

1. x < 0, v < 0, a > 0
2. x > 0, v < 0, a < 0
3. x > 0, v < 0, a > 0
4. x > 0, v > 0, a > 0

Answer:

As the lift is coming down, the value of x becomes less hence negative, i.e., x < 0.

Velocity is downwards (i.e., negative). So v < 0.

Just before reaching the ground floor, the lift is retarded, i.e., acceleration is upwards. Hence a > 0.

We can conclude that x < 0, v < 0 and a > 0.

∴ (1) is correct.

Question 6. A uniformly moving cricket ball is hit with a bat for a very short time and is turned back. Show the variation of its acceleration with time taking the acceleration in the backward direction as positive.
Answer:

For a ball moving with uniform velocity acceleration is zero. But during the time of contact between the ball and bat acceleration is applied in the opposite direction. The shape of the acceleration – time graph is as shown.

Question 7. Give an example of one-dimensional motion where a particle moving along the positive x-direction comes to rest periodically and moves forward.
Answer:

When the length of the pendulum is high and the amplitude is less then its motion is along a straight line. The pendulum will come to a stop at the extreme position and move back in the forward direction (‘x’ + ve) periodically.

Question 8. An object falling through a fluid is observed to have an acceleration given by a = g – bv where g is the gravitational acceleration and b is a constant. After a long time, it is observed to fall with a constant velocity. What would be the value of this constant velocity?
Answer:

Acceleration, a = g – bv when moving with constant velocity, a = 0 ⇒ 0 = g – bv

∴ Constant velocity, v = \(\frac{g}{b}\) m/sec.

Question 9. If the trajectory of a body is parabolic in one frame, can it be parabolic in another frame that moves with a constant velocity with respect to the first frame? If not, what can it be?
Answer:

If the trajectory of a body is parabolic with reference frames one and two then those two frames are of rest or moving with uniform velocity.

If they are not parabolic then for that reference frame it may be in a straight line path.

Example: When a body is dropped from a moving plane its path is parabolic for a person outside the plane. But for the pilot in the plane, it is falling vertically downwards.

Question 10. A spring with one end attached to amass and the other to rigid support is stretched and released. When is the magnitude of acceleration a maximum?
Answer:

Maximum restoring force set up in the spring when stretched by a distance ‘r’ is F = – kr

Potential energy of stretched spring = \(\frac{1}{2}\) kx²

As F ∝ r and this force are directed towards the equilibrium position, hence if the mass is left free, it will execute damped SHM due to gravity pull.

Magnitude of acceleration in the mass attached to one end of spring when just released is \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{-\mathrm{k}}{\mathrm{m}} \mathrm{r}=(\text { Maximum })\)

∴ The magnitude of the spring’s acceleration will be maximum when it is just released.

KSEEB Class 11 Physics Chapter 3 Solutions

Question 11. Define average velocity and average speed. When does the magnitude of average velocity become equal to the average speed?
Answer:

Average Velocity: It is defined as the ratio of total displacement to total time taken.

Average velocity = \(\frac{\text { total displacement }}{\text { total time taken }}=\frac{\mathrm{x}_2-\mathrm{x}_1}{\mathrm{t}_2-\mathrm{t}_1}\)

Average velocity is independent of the path followed by the particle. It just deals with the initial and final positions of the body.

Average Speed: The ratio of the total path length travelled to the total time taken is known as “average speed”.

Speed and average speed are scalar quantities so no direction for these quantities.

Average speed = \(\frac{\text { Total path length }}{\text { Total time interval }}\)

When the body is along with the straight line its average velocity and average speed are equal.

Solutions For Motion In A Straight Line KSEEB Physics Chapter 3 Short Answer Questions

Question 1. Can the equations of kinematics be used when the acceleration varies with time? If not, what form would these equations take?
Answer:

The equations of motion are

1. v =u + at
2. s = ut + \(\frac{1}{2}\)at² and
3. v² – u² = 2as.

All these three equations are applicable to body moves with uniform acceleration ‘a’.

No, the equations are not applicable when the acceleration varies with time.

Question 2. A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v₁ and at time t₂ = t is v₂. The average velocity of the particle in this time interval is (v₁ + v₂) / 2. Is this correct? Substantiate your answer.
Answer:

t₁ = 0 ⇒ u = v₁; t₂ = t ⇒ v = v₂

Average velocity = \(\frac{u+v}{2}=\frac{v_1+v_2}{2}\)

Question 3. Can the velocity of an object be in a direction other than the direction of acceleration of the object? If so, give an example.
Answer:

Yes, the Velocity of a body and its acceleration may be in different directions.

Explanation:

1. In the case of a vertically projected body ⇒ velocity of the body is in the upward direction and acceleration is in a downward direction.
2. When brakes are applied the velocity of the body before coming to rest is opposite to retarding acceleration.

Question 4. A parachutist flying in an aeroplane jumps when it is at a height of 3 km above the ground. He opens his parachute when he is about 1 km above ground. Describe his motion.
Answer:

Initially, the path is a parabola as seen by an observer on the ground. It is a vertical straight line as seen by the pilot. He opens his parachute, it is moving vertically downwards with decreasing velocity and finally, it reaches the ground.

Question 5. A bird holds n fruit In its beak and flics parallel to the ground. It lets go of the fruit at some height. Describe the trajectory of the fruit as it falls to the ground as seen by

1. The bird
2. A person on the ground.

Answer:

1. As the bird Is flying parallel to the ground, it possesses velocity in the horizontal direction. Hence the fruit also possesses velocity in the horizontal direction and acceleration in a downward direction. Hence the path of the fruit is a straight line with respect to the bird.
2. With respect to a person on the ground, the fruit seems to be on a parabolic path.

Question 6. A man runs across the roof of a tall building and jumps horizontally onto the (lower) roof of an adjacent building. If his speed is 9 ms^-1 the horizontal distance between the buildings is 10 m and the height difference between the roofs is 9 m, will he be able to land on the next building? (take g = 10 ms^-2)
Answer:

Given that,

initial speed, u = 9 ms^-1 ; g = 10m/s²

The height difference between the roofs, h = 9 m

Horizontal distance between two buildings, d = 10 m

Time of flight \(\mathrm{T}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 9}{10}}\)

= \(\sqrt{1.8}=1.341 \mathrm{~s}\)

Range of the man = R = u x T = 9 x 1.341 s = 12.069 m

Since R > d, the man will be able to land on the next building.

Question 7. A ball is dropped from the roof of a tall building and simultaneously another ball Is thrown horizontally with some velocity from the same roof. Which ball lands first? Explain your answer.
Answer:

Let ‘h’ be the height of the tall building.

For Dropped Ball: Let ‘t₁‘ be the time taken by the dropped ball to reach the ground.

Initial velocity, u = 0 ; Acceleration, a = + g

Distance travelled, s = h; Time of travel, t = t₁

From the equation of motion, s = ut + \(\frac{1}{2}\) at²

We can write, h = \(0 \times t_1+\frac{1}{2} \times g \times t_1^2 \Rightarrow h=\frac{1}{2} g t_1^2 \Rightarrow t_1^2=\frac{2 h}{g}\)

(or) \(t_1=\sqrt{\frac{2 h}{g}}\)…..(1)

For Horizontally Projected Ball: If the ball is thrown horizontally then its initial velocity along the vertical direction is zero and in this case, let ‘t₂‘ be the time taken by the ball to reach the ground.

Again from the equation of motion,

s = \(\mathrm{ut}+\frac{1}{2} \mathrm{at}^2\) we can write,

⇒ h = \(0 \times \mathrm{t}_2+\frac{1}{2} \mathrm{~g} \times \mathrm{t}_2^2\)

⇒ h = \(\frac{1}{2} \mathrm{gt}_2^2 ;\)

⇒ \(\mathrm{t}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~g}}\) (or)

∴ \(\mathrm{t}_2=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)…….(2)

From equation (1) and (2) t₁ = t₂

i.e., both the balls reach the ground at the same time.

Question 8. A ball Is dropped from a building and simultaneously another ball is projected upward with some velocity. Describe the change In relative velocities of the balls as a function of time.
Answer:

1. For a body dropped from building its velocity, v₁ = gt…..(1) (u₁ = 0)
2. For a body thrown up with a velocity ‘u’ its velocity, v² = u – gt….(2)

The two balls are moving in the opposite direction the relative velocity, v_R = v₁ + v₂

∴ v_R = gt + u – gt = u

Here the relative velocity remains constant, but the velocity of one body increases at a rate of ‘g’ m/sec and the velocity of another body decreases at a rate of ‘g’m/sec.

Question 9. A typical raindrop is about 4 mm in diameter. If a raindrop falls from a cloud which is 1 km above the ground, estimate its momentum when it hits the ground.
Answer:

Diameter, D = 4 m

⇒ radius, r = 2mm = 2 x 10^-3 m

mass of raindrop = volume x density = \(\frac{4}{3} \pi r^3 \times 1000 \mathrm{~m}\)

(mass of one m³ of water = 1000 kg)

∴ m = \(\frac{4}{3} \times \frac{22}{7} \times\left[2 \times 10^{-3}\right]^3 \times 1000\)

= \(33.52 \times 10^{-6} \mathrm{~kg}\)

Velocity of raindrop \(v=\sqrt{2 g h}\)

But h = 1 km = 1000 m

∴ \(\mathrm{v}=\sqrt{2 \times 9.8 \times 1000}=\sqrt{19600}\) = 140 m/sec

Momentum of raindrop, \(\vec{p}=m v\)

= 33.52 x 10^-6 x 140 = 4.693 x 10^-3 kg-m

Question 10. Show that the maximum height reached by a projectile launched at an angle of 45° is one-quarter of its range.
Answer:

In projectiles Range, \((R)=\frac{u^2 \sin 2 \theta}{g}\)

Maximum height \((H)=\frac{u^2 \sin ^2 \theta}{2 g}\)

When \(\theta=45^{\circ}\)

⇒ \(\mathrm{R}_1=\mathrm{R}_{\max }=\frac{\mathrm{u}^2 \sin 90^{\circ}}{\mathrm{g}}\)

∴ \(\mathrm{R}_{\max }=\frac{\mathrm{u}^2}{\mathrm{~g}}\)

Maximum height \(H_{\max }=\frac{u^2 \sin ^2\left(45^{\circ}\right)}{2 g}\)

= \(\frac{u^2(1 / \sqrt{2})^2}{2 g}=\frac{u^2}{4 g}\)

∴ \(H_{\max }=\frac{1}{4} R_{\max }\)

∴ When θ = 45° maximum height reached is one-quarter of the maximum range.

Question 11. Derive the equation of motion x = v₀t + \(\frac{1}{2}\) at² using appropriate graph.
Answer:

The velocity-time graph of a body moving with initial velocity ‘u’ and with uniform acceleration ‘a is shown. Let V be the velocity of the body after a time t.

In the v -t graph area of the velocity-time graph = total displacement travelled by it. The area under velocity – time graph = area of OABCD

∴ Area of Rectangular part OACD = Area of OACD + Area of ABC.

A₁ = OA x OD = v₀ t…….(1)

Area of triangle ABC = A₂

A₂ = \(\frac{1}{2}\) Base x height

= \(\frac{1}{2}\) AC x (BC)

= \(\frac{1}{2}\)t(v-v)

But v – v₀ = at

∴ A₂ = \(\frac{1}{2}\)t at = \(\frac{1}{2}\) at²

∴ Total area under graph = s = A₁ + A₂

s(n)= \(v_0 t+\frac{1}{2} a t^2\) is graphically proved.

KSEEB Physics Class 11 Motion in a Straight Line Chapter 3 Problems

Question 1. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home at a speed of 7.5 km h^-1. What is the

1. Magnitude of average velocity and
2. The average speed of the man over the time interval of 0 to 50 minutes?

Solution:

Time taken by man to go from his home to market, \(\mathrm{t}_1=\frac{\text { distance }}{\text { speed }}=\frac{2.5}{7.5}=\frac{1}{3} \mathrm{~h}\)

∴ Total time taken = \(t_1+t_2=\frac{1}{2}+\frac{1}{3}=\frac{5}{6} h\) = 50 min

In time interval 0 to 50 min

Total distance travelled = 2.5 + 2.5 = 5 km.

Total displacement = zero.

1. Average velocity = \(\frac{\text { displacement }}{\text { time }}=0\)
2. Average speed = \(\frac{\text { displacement }}{\text { time }}=\frac{5}{\frac{5}{6}}\) = 6 km/h

Question 2. A stone is dropped from a height of 300 m and at the same time, another stone is projected vertically upwards with a velocity of 100 m/sec. Find when and where the] two stones meet.
Solution:

Height h = 300 m ;

Initial velocity u₀ = 100 m/s

Let the two stones meet at a height ‘x’ above the ground.

For 1st stone \(\mathrm{h}-\mathrm{x}=\frac{1}{2} \mathrm{gt}^2\)…..(1)

For 2nd stone \(\mathrm{x}=\mathrm{u}_0 \mathrm{t}-\frac{1}{2} \mathrm{gt}^2\)

⇒ \(\frac{1}{2} g t^2=u_0 t-x\)….(2)

Since t is the same for the two stones

From equations 1 and 2.

h – x = u₀t – x

⇒ \(\mathrm{u}_0 \mathrm{t}=\mathrm{h} \text { or time } \mathrm{t}=\frac{\mathrm{h}}{\mathrm{u}_{\mathrm{o}}}=\frac{300}{100}=3 \mathrm{sec}\).

⇔ The two stones will meet 3 seconds after the 1st stone is dropped or the 2nd stone is thrown up.

Question 3. A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph. What is its mean speed over the entire distance?
Solution:

Total distance = s

distance travelled, \(\mathbf{s}_1=\frac{\mathbf{s}}{3}\)

velocity, v₁ = 10 kmph

distance, \(\mathbf{s}_2=\frac{\mathbf{s}}{3}\)

velocity, v₂ = 20 kmph

distance, \(\mathbf{s}_3=\frac{\mathbf{s}}{3}\)

velocity, v₃ = 60 kmph

Average velocity = \(\frac{\text { total distance }}{\text { total time }}\)

Time, \(t=t_1+t_2+t_3=\frac{s}{3 v_1}+\frac{s}{3 v_2}+\frac{s}{3 v_3}\)

(because time = \(\frac{\text { distance }}{\text { velocity }}\)

∴ \(\mathrm{t}=\frac{\mathrm{s}}{3}\left(\frac{\mathrm{v}_2 \mathrm{v}_3+\mathrm{v}_3 \mathrm{v}_1+\mathrm{v}_1 \mathrm{v}_2}{\mathrm{v}_1 \mathrm{v}_2 \mathrm{v}_3}\right)\)

= \(\frac{\mathrm{s}}{3}\left[\frac{(60 \times 20)+(60 \times 10)+(20 \times 10)}{60 \times 20 \times 10}\right]\)

= \(\frac{\mathrm{s}}{3} \frac{(1200+600+200)}{12000}\)

= \(\frac{\mathrm{s}}{3} \frac{2000}{12000}=\frac{2000 \mathrm{~s}}{3 \times 12000}\) (Or)

∴ \(v_{\text {mean }}==\frac{3 v_1 v_2 v_3}{v_1 v_2+v_2 v_3+v_3 v_1}=18 \mathrm{kmph}\)

Average velocity  = \(\frac{\mathrm{s}}{\mathrm{t}}\)

= \(\frac{3 \times 12000}{2000 \mathrm{~s}}\) = 18 kmph

Question 4. A bullet moving with a speed of 150 m s^-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree?
Solution:

Velocity of bullet, u = 150 m/s;

Final velocity, v = 0

Distance travelled, s = 3.5 cm = 3.5 x 10^-2 m,

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

= \(\frac{0^2-150^2}{2 \times 3.50 \times 10^{-2}}\)

= \(\frac{-22500}{7 \times 10^{-2}}\)

= \(-3.214 \times 10^5 \mathrm{~m} / \mathrm{sec}^2\) (-ve sign for retardation)

Time taken to stop, \(\mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\)

= \(\frac{-150}{-3.214 \times 10^5}\)

= \(4.67 \times 10^{-4} \mathrm{sec}\)

Motion In A Straight Line Problems in KSEEB Physics

Question 5. A motorist drives north for 30 min at 85 km/h and then stops for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity?
Solution:

In First Part:

Velocity, v₁ = 85 kmph

Time, t₁ = 30 min

Distance travelled, s₁ = v₁ t₁

= 85 x \(\frac{30}{60}\) =42.5 km

In Second Part:

Distance travelled, s₂ = 0;

Time, t₂ = 15.0 min.

In Third Part:

Distance travelled, s₃ = 130 km;

Time, t₃ = 120 min = 2 hours

1. Total distance of the motorist, s = s₁+ s₂ + s₃ = 42.5 + 0 + 130 = 172.5 km

2. Total time travelled, t = t₁ + t₂ + t₃ = 30 + 15 + 120

= 165 minutes

= 2 hrs 45 minutes

= 2\(\frac{3}{4}\) hrs. = \(\frac{11}{4}\) hrs.

∴ Average velocity, \(\mathrm{v}_{\mathrm{avg}}=\frac{\text { total displacement }}{\text { total time }}\)

= \(\frac{172.5}{(11 / 4)}=62.7 \mathrm{kmph}\)

Question 6. A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collision occur?
Solution:

Given at time of collision velocity of A = V_A

= 2 x V_B (velocity of B)

Let the body be dropped from a height ‘h’.

Let the two stones collide at x from the ground.

For the body dropped, \(\mathrm{s}=\mathrm{h}-\mathrm{x}=\frac{1}{2} \mathrm{gt}^2\)……(1)

For the body thrown up, x = ut – \(\frac{1}{2}\) gt²…….(2)

For the body dropped, v = u + at ⇒ V_A = gt…….(3)

For the body thrown up, v = u – gt ⇒ V_B = u – gt……..(4)

Given V_A = 2V_B

⇒ gt = 2 (u – gt) or u = \(\frac{3gt}{2}\)…… (5)

Divide equation (1) with equation (2)

⇒ \(\frac{h-x}{x}=\frac{\frac{1}{2} g t^2}{u t-\frac{1}{2} g t^2} \Rightarrow \frac{h}{x}-1=\frac{\frac{1}{2} g t^2}{u t-\frac{1}{2} g t^2}\)

⇒ \(\frac{h}{x}=\frac{\frac{1}{2} g t^2+u t-\frac{1}{2} g t^2}{u t-\frac{1}{2} g t^2}\)

∴ \(\frac{h}{x}=\frac{u t}{u t-\frac{1}{2} g t^2}\)

Put u = \(\frac{3 g t}{2}\) then \(\frac{h}{x}=\frac{3 g t^2 / 2}{3 g t^2 / 2-\frac{1}{2} g t^2}\)

∴ \(\frac{h}{x}=\frac{3 g t^2}{2 g t^2}=\frac{3}{2}\)

∴ x = \(\frac{2}{3} h\)

∴ Fraction of height of collision = \(\frac{2}{3}\)

KSEEB Class 11 Physics Motion in a Straight Line Key Concepts

Question 7. Drops of water fall at regular intervals from the roof of a building of height 16 m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Solution:

Height of building, h = 16 m

∴ Time taken to reach the ground, t = \(\sqrt{\frac{2 h}{g}}\)

∴ Time taken to fall, t =\(\sqrt{\frac{2 \times 16}{9.8}}\)

= \(\sqrt{\frac{32}{9.8}}=\sqrt{3.26}=1.80\)

Number of drops, n = 5

∴ Number of intervals = n — 1 = 5 — 1 = 4

Time interval between drops = \(\frac{1.8}{4}\) = 0.45 sec

Time of travel of 1st drop, t₁ = 4 x 0.45 = 1.81

∴ Distance travelled by 1st drop, \(s_1=\frac{1}{2} g t_1^2=\frac{1}{2} \times 9.8 \times 1.8 \times 1.8=16 \mathrm{~m}\)

For 2nd drop, t₂ = 3 x0.45 = 1.35 sec., \(\mathrm{S}_2=\frac{1}{2} \mathrm{gt}_2^2\);

= \(4.9 \times 1.822 \simeq 1.822=9 \mathrm{~m}\)

For 3rd drop, \(\mathrm{t}_3=2 \times 0.45=0.9 \mathrm{sec}\)

Distance \(\mathrm{S}_3=\frac{1}{2} \mathrm{gt}_3^2=\frac{1}{2} \times 9.8 \times 0.9^2 =3.97\) = 4m

For 4th drop, \(t_4=1 \times 0.45=0.45 \mathrm{sec}\)

Distance travelled, \(\mathrm{S}_4=\frac{1}{2} \mathrm{gt}_4^2=\frac{1}{2} \times 9.8\) \(\times(0.45)^2 \simeq 1\)

For 5 th drop, t₅ = 0 ⇒ S₅ = 0

Distance between 1st and 2nd drop S_1,2 = S₁ – S₂ = 16-9 = 7 m

Distance between 2nd and 3rd drop S_2,3 = S₂ – S₃ = 9 – 4 = 5m

Distance between 3rd and 4th drop S_3,4 = S₃ – S₄ = 4 – 1= 3 m

Distance between 4th and 5th drop S_4,5 = S₄ – S₅ = 1-0 = 1 m

∴ Distances between successive drops are 7m, 5m, 3m and lm.

Question 8. Rain is falling vertically with a speed of 35 ms^-1. A woman rides a bicycle with a speed of 12 ms^-1 in east to west direction. What is the direction in which she should hold her umbrella?
Answer:

Velocity of rain V_R = 35 m/s (vertically)

Velocity of women V_w = 12 m/s (towards east)

Resultant angle = \(\theta=\tan ^{-1} \frac{\mathrm{V}_{\mathrm{W}}}{\mathrm{V}_{\mathrm{R}}}=\frac{12}{35}\)

∴ \(\theta=\tan ^{-1}\left[\frac{12}{35}\right]=0.343 \text {. or } \theta=19^{\circ}\) (Nearly)

She should hold an umbrella at an angle of 19° with the east.

Question 9. A hunter aims a gun at a monkey hanging from a tree some distance away. The monkey drops from the branch at the moment he fires the gun hoping to avoid the builet. Explain why the monkey made a wrong move.
Answer:

Let the bullet be fired with an angle α and distance from hunter’s rifle to monkey = x

The vertical component of velocity v_xy = v sin α

when exactly aimed at monkey S_y = v sinα t = h

But acceleration due to gravity \(h_1=u \sin \alpha \mathrm{t}-\frac{1}{2} g t^2=h-\frac{1}{2} g t^2\)….(1)

So the bullet passes through a height of \(\frac{1}{2}\) gt² below the monkey.

But when the monkey is falling freely height  of fall during time t = \(\frac{1}{2}\)

So new height is h – \(\frac{1}{2}\) gt² …… (2)

From equations (1) and (2) h₁ is the same i.e., if the monkey is dropped from the branch bullet will hit it exactly.

KSEEB Physics Motion In A Straight Line Questions And Answers

Question 10. A food packet is dropped from an aeroplane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500m. Find

1. It is a time of descent
2. The horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.

Solution:

The velocity of plane, V = 360 kmph = 360 x \(\frac{5}{18}\) = 100 m/s

Height above ground, h = 500 m; g = 10 m/s²

1. Time of descent, \(\mathrm{t}=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}=\sqrt{100}=10 \mathrm{sec}\)

2. Horizontal distance between the point of dropping and the point where it reaches the ground = Range R.

∴ R = \(u \sqrt{\frac{2 h}{g}}=100 \sqrt{\frac{2 \times 500}{10}}\) = 100 x 10 = 1000 m.

Question 11. A ball is tossed from the window of a building with an initial velocity of 8 ms^-1 at an angle of 20° below the horizontal. It strikes the ground 3 s later. From what height was the ball thrown? How far from the base of the building does the ball strike the ground?
Solution:

Initial velocity, u = 8 m/s;

The angle of projection, θ = 20°

Time is taken to reach the ground, t = 3 sec.

The horizontal component of initial velocity, u_x = u cos θ = 8 cos 20° = 8 x 0.94 = 7.52 m/s

Vertical component of initial velocity, v_y = u sin θ = 8 sin 20° = 8 x 0.342 = 2.736 m/s

1. From equation of motion, s = ut + \(\frac{1}{2}\) at²

we can write h = (u sinθ)t + \(\frac{1}{2}\) gt²

⇒ h = (2.736)3 + \(\frac{1}{2}\) x 9.8 x(3)² .

⇒ h = 8.208+ 4.9 x 9

⇒ h = 8.208 + 44.1 or h = 52.308 m

2. Horizontal distance travelled, s_x = v_x x t = 7.52 x 3 = 22.56 m

Question 12. Two balls are projected from the same point in directions 30° and 60° with respect to the horizontal. What is the ratio of their initial velocities if they

1. Attain the same height?
2. Have the same range?

Solution:

The angle of projection of the first ball, θ₁ =30°

The angle of projection of the second ball, θ₂ = 60°

Let u₁ and u₂ be the velocities of projections of the two balls.

1. Maximum height of the first ball, \(\left(\mathrm{H}_{\max }\right)_1=\frac{\mathrm{u}_1^2 \sin ^2 30^{\circ}}{2 \mathrm{~g}}\)

Maximum height of second ball, \(\left(\mathrm{H}_{\max }\right)_2=\frac{\mathrm{u}_2^2 \sin ^2 60^{\circ}}{2 \mathrm{~g}}\)

Given that \(\left(H_{\max }\right)_1=\left(H_{\text {max }}\right)_2\)

⇒ \(\frac{u_1^2 \sin ^2 30^{\circ}}{2 g}=\frac{u_2^2 \sin ^2 60^{\circ}}{2 g}\)

⇒ \(u_1^2 \sin ^2 30^{\circ}=u_2^2 \sin ^2 60^{\circ}\)

⇒ \(\frac{u_1^2}{u_2^2}=\frac{\sin ^2 60^{\circ}}{\sin ^2 30^{\circ}} \Rightarrow \frac{u_1^2}{u_2^2}=\frac{(\sqrt{3} / 2)^2}{(1 / 2)^2}\)

⇒ \(\frac{u_1}{u_2}=\frac{\sqrt{3}}{1}\) or \(u_1: u_2=\sqrt{3}: 1\)

2. If the balls have the same range, then \(R_1=R_2\)

i.e., \(\frac{u_1^2 \sin \left(2 \times 30^{\circ}\right)}{g}=\frac{u_2^2 \sin \left(2 \times 60^{\circ}\right)}{g}\)

⇒ \(u_1^2 \sin 60^{\circ}=u_2^2 \sin 120^{\circ}\)

⇒ \(\frac{u_1^2}{u_2^2}=\frac{\sin 120^{\circ}}{\sin 60^{\circ}} \Rightarrow \frac{u_1^2}{u_2^2}=\frac{\cos 30^{\circ}}{\sin 60^{\circ}}\)

⇒ \(\frac{u_1^2}{u_2^2}=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} \text { or } u_1: u_2=1: 1\)

Question 13. A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a multistory building. The height of the point from where the hall Is thrown Is 25.0 in from the ground.

1. How high will the ball rise?
2. How long will it be before the ball hits the ground?
3. Take g = 10 ms^-2 [Actual value of ‘g’ is 9.8 ms^-2]

Solution:

Initial velocity V₀ = 20 m/s;

Height above ground h₀ = 2.50 m ; g = 10 m/s²

1. For a body thrown up vertically height of rise \(\left(y-y_0\right)=\frac{u^2}{2 g}=\frac{20 \times 20}{2 \times 10}=20 \mathrm{~m}\)

2. Time spent in the air (t) is \(y_1-y_0=V_0 t+\frac{1}{2} g t^2\)

Where y₁ = Total displacement of the body from the ground = 0

∴ 0 = \(y_0+V_0 t+\frac{1}{2} g t^2=25+20 t-\frac{1}{2} \cdot 10 . t^2\)

(because \(g=-10 \mathrm{~m} / \mathrm{s}^2\)) while going up

∴ \(0=-5 t^2+20 t+25\) (or) \(t^2-4 t-5=0\)

i.e., (t-5)(t+1)=0

t=5 (or) t=-1

But time is not – ve.

∴ Time spent in air t = 5 sec

KSEEB Class 11 Physics Chapter 3 Motion In A Straight Line 

Question 14. A parachutist flying in an aeroplane jumps when it is at a height of 3 km above ground. He opens his parachute when he is about 1 km above ground. Describe his motion.
Answer:

1. Height of fall before opening, h = 2 km = 2000 m

∴ Velocity at a height of 1 km

= \(\sqrt{2 \mathrm{gh}_1}=\sqrt{2 \times 10 \times 2000}\)

= \(\sqrt{40000}=200 \mathrm{~ms}^{-1}\)

Time of fall, t = \(\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 2000}{10}}\)

= \(\sqrt{400}=20 \mathrm{sec}\)

2. After the parachute Is opened It touches the ground with almost zero velocity.

∴ u = 200 m/sec, v = 0, S = h = 1000 m

From v² – u² = 2as

Retardation, \(a=\frac{0-200^2}{1000}=-40 \mathrm{~m} / \mathrm{sec}^2\)

Time taken to reach ground, t = \(\frac{u}{a}\) = \(\frac{200}{40}\) = 5 sec

The month is as shown.

Question 15. A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?
Solution:

Initial speed = V₀ = 126 kmph

= 126 x \(\frac{5}{18}\) = 35 m/sec

Ending speed = V = 0

Distance taken to stop = X = 200 m.

From V² – V₀² = 2ax .

a = \(\frac{-35}{2 \times 200}\) = \(\frac{-35 \times 35}{2 \times 200}\) = \(\frac{-49}{16}\)

Time taken to stop t =?

From V = V₀ + at

0 = \(35+\frac{49}{16} \times t \Rightarrow t=\frac{35 \times 6}{49}\) = \(\frac{80}{7}=11.43 \mathrm{sec}\)

Units and Measurements Solutions KSEEB Class 11 Physics Chapter 2 Very Short Answer Questions

Question 1. Distinguish between accuracy and precision.
Answer:

Accuracy:

1. It is defined as the closeness of the measured value to the true value.
2. It depends on the minimization of errors.

Precision:

1. It is defined as to what resolution the quantity is measured.
2. It depends on the least count of the measuring instrument.

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. What are the different types of errors that can occur in a measurement?
Answer:

Types Of Errors

1. Systematic errors and
2. Random errors.

Systematic Errors Are Again Divided Into

1. Imperfection errors
2. Environmental errors and
3. Personal errors.

KSEEB Class 11 Physics Chapter 2

Question 3. How can systematic errors be minimized or eliminated?
Answer:

Systematic Errors Can Be Minimised

1. By improving experimental techniques,
2. By selecting better instruments.

Question 4. Illustrate how the result of a measurement is to be reported indicating the error involved.
Answer:

The result of the measurement is in the form of a number which indicates the precision of measurement along with the unit of the same physical quantity.

Question 5. What are significant figures and what do they represent when reporting the result
of a measurement?
Answer:

Significant figures represent all practically measured digits plus one uncertain digit at the end.

When a result Is reported in this way we can know up to what extent the value Is reliable and also the amount of uncertainty In that reported value.

Question 6. Distinguish between fundamental units| and derived units.
Answer:

The units of fundamental physical| quantities are called fundamental units.

Example: Kg, m, sec, etc.,

The units of derived physical quantities are called derived units.

Example: m/sec, J, m/sec² etc.,

Question 7. Why do we have different units for the same physical quantity?
Answer:

To measure the same physical quantity we have different units by keeping the magnitude of the quantity to be measured.

Example:

1. The measure astronomical distances we will use light year.
   • 1 light year = 9.468 x 10¹⁵ m.
2. To measure atomic distances we will use Angstrom Å (or) Fermi.

Question 8. What is dimensional analysis?
Answer:

Dimensional analysis is a tool to check the relations among physical quantities by using their dimensions.

Dimensional analysis is generally used to check the correctness of derived equations.

KSEEB Class 11 Physics Units And Measurements Notes And Solutions

Question 9. How many orders of magnitude greater is the radius of the atom than that of the nucleus?
Answer:

Size of atom = 10^-10 m,

Size of atomic nucleus = 10^-14 m.

Size of the atom + size of the nucleus is \(\frac{10^{-10}}{10^{-14}}=10^4\)

∴ The size of an atom is 10⁴ times greater than the size of the nucleus.

Question 10. Express unified atomic mass unit in kg.
Answer:

By definition,

1 a.m.u. = \(\frac{1}{12} \times \text { mass of an atom of }{ }_6^{12} \mathrm{C}\)

= \(\frac{1}{12} \times \frac{12 \mathrm{~g}}{6.023 \times 10^{23}}\)

= \(1.67 \times 10^{-24} \mathrm{~g}=1.67 \times 10^{-27} \mathrm{~kg}\)

Chapter 2 Units And Measurements Short Answer Questions

Question 1. The vernier scale of an instrument has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then using this instrument what would be the minimum inaccuracy in the measurement of distance?
Answer:

Least count of Vernier Callipers = 1 MSD – 1 VSD

∴ L.C. = 1 MSD – \(\frac{49}{50}\) MSD = \(\frac{1}{50}\) MSD

= \(\frac{1}{50}\) x 0.5 = 0.01 m.m

KSEEB Physics Class 11 Units and Measurements Key Concepts

Question 2. In a system of units, the unit of force is 100N, the unit of length is 10m and the unit of time is 100s. What is the unit of mass in this system?
Answer:

Here, F = MLT^-2 = 100 N …….(1) ;

L = 10 m ; T = 100s

∴ From equation (1)

M X (10) X (100)^-2 = 100

⇒ M X 10^-3 = 100

⇒ M = \(\frac{100}{10_3}\)

⇒ M = 10⁵ kg

Question 3. The distance of a galaxy from Earth is of the. order of 10²⁵ m. Calculate the order of magnitude of the time taken by light to reach us from the galaxy.
Answer:

Size of galaxy = 10²⁵ m,

Velocity of light, c = 3 x 10⁸ ms^-1

Time taken by light to reach Earth,

t = \(\frac{d}{c}=\frac{10^{25}}{3 \times 10^8}=\frac{1}{3} \times 10^{17}=0.3333 \times 10^{17}\)

While calculating the order of magnitude we will consider the power of Ten only.

So the order of magnitude of time taken by light to reach Earth from the galaxy is 10¹⁷ seconds.

Question 4. The Earth-Moon distance is about 60 Earth radius. What will be the approximate diameter of the Earth as seen from the Moon?
Answer:

Earth-moon distance, D = 60r.

Diameter of earth, b = 2r

As seen from the moon

⇒ \(\theta=\frac{\text{Arc~} A B}{\text{Rad}}=\frac{2 \mathrm{r}}{60 \mathrm{r}}=\frac{1}{30} \mathrm{Rad}\)

∴ The size of the earth as seen from the moon is \(\frac{1}{30}\)

radians, or \(\frac{57^{\circ} 30^{\prime}}{30} \cong 1.9^{\circ} \text { (or) } 2^{\circ}\)

Units And Measurements Questions And Answers KSEEB Physics

Question 5. Three measurements of the time for 20 oscillations of a pendulum give t₁ = 39.6 s, t₂ = 39.9 s, and t₃ = 39.5 s. What is the precision of the measurements? What is the accuracy of the measurements?
Answer:

Precision is the least measurable value with that instrument in our case precision is = 0.1 sec.

Calculation Of Accuracy:

Average value of measurements = \(\frac{39.6+39.9+39.5}{3}=\frac{119}{3}=39.67\)

Error in each measurement

⇒ \(\Delta a_1=39.6-39.6=0.07\);

⇒ \(\Delta a_2=39.9-39.67=0.23\);

⇒ \(\Delta a_3=39.67-39.5=0.17\)

∴ \(\Delta a_{\text {mesn }}=\frac{0.07+0.23+0.17}{3}=0.156\)

Precision ± 1 sec. In these measurements only two significant figures are believable. 3rd one is uncertain.

Adjustment of Δa_mean up to the given significant figure = 0.156 adjusted to 0.2.

So our value is accurate up to ±0.2

So our result is 39.67 ± 0.2, when significant figures are taken into account it is 39.7 ± 0.2 sec.

Question 6. 1 calorie = 4.2 J where 1 J = 1 kg m²s^-2. Suppose we employ a system of units in which the unit of mass is α kg, the unit of length is β m, and the unit of time γ s, showing that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in the new system.
Answer:

Here, 1 calorie = 4.2 J = 4.2 kg m²/s² ……(1)

As a new unit of mass = α kg

∴ 1 kg = \(\frac{1}{\alpha}\) new unit of mass

⇒ α^-1 new unit of mass

Similarly 1 m = β^-1 new unit of length and 1 s = γ^-1 new unit of time

Putting these values in (1) we get

1 calorie = 4.2 (α^-1 new unit of mass) (β^-1 new unit of length)² (γ^-1  new unit of time)^-2

= 4.2 α^-1β^-2γ² new unit of energy, which was proved.

Question 7. A new unit of length is chosen so that the speed of light in vacuum is 1 ms^-2. If light takes 8 min and 20s to cover this distance, what is the distance between the Sun and Earth in terms of the new unit?
Answer:

Given the velocity of light in a vacuum,

c = 1 new unit of length s^-1

Time taken by the light of the Sun to reach Earth,

t = 8 min 20s = 8 x 60 + 20 = 500 s

∴ Distance between the Sun and Earth, x = c x t

= 1 new unit of length s^-1 x 500s

= 500 new units of length.

KSEEB Class 11 Physics Chapter 2 Units and Measurements Solutions

Question 8. A student measures the thickness of a human hair using a magnification 100 microscope. He makes 20 observations and finds that the average thickness (as viewed in the microscope) is 3.5 mm. What is the estimate of the thickness of hair?
Answer:

Magnification, \(\mathrm{m}=\frac{\text { observed width }(\mathrm{y})}{\text { real width }(\mathrm{x})}\)

⇒ x = \(\frac{\mathrm{y}}{\mathrm{m}}=\frac{3.5 \mathrm{~mm}}{100}=0.035 \mathrm{~mm}\)

∴ The thickness of the hair is 0.035 mm

Question 9. A physical quantity X is related to four mea¬surable quantities a, b, c, and d as follows?

X = a²b³C^5/2d^-2

The percentage error in the measurement of a, b, c, and d are 1%, 2%, 3% and 4% respectively. What is the percentage error in X?

Answer:

Here, X = a²b³C^5/2d^-2

⇒ \(\frac{\Delta X}{X}= \pm\left[2\left(\frac{\Delta a}{a}\right)+3\left(\frac{\Delta b}{b}\right)+\frac{5}{2}\left(\frac{\Delta c}{c}\right)+2\left(\frac{\Delta d}{d}\right)\right] \)

= \(\pm\left[2(1 \%)+3(2 \%)+\frac{5}{2}(3 \%)+2(4 \%)\right]\)

= ± 423.5 %

The percentage error in X is ± 23.5 %

Question 10. The velocity of a body is given by v = At² + Bt + C. If v and t are expressed in SI what are the units of A, B, and C?
Answer:

From the principle of Homogeneity, the terms At², Bt, and C must have the same dimensional formula of velocity V.

v = \(\text { Velocity }=\mathrm{LT}^{-1} \Rightarrow \mathrm{LT}^{-1}=\mathrm{A}\left[\mathrm{T}^2\right]\)

∴ A = \(\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^2}=\mathrm{LT}^{-3}\) So unit of A is m/\(\mathrm{sec}^3\)

∴ \(\mathrm{LT}^{-1}=\mathrm{BT} \Rightarrow \mathrm{B}=\mathrm{LT}^{-2}\) So unit of B is \(\mathrm{m} / \mathrm{sec}^2\)

∴ \(\mathrm{LT}^{-1}=\mathrm{C}\)

So the unit of C is \(\mathrm{m} / \mathrm{sec}\).

Dimensional Formulae Of Physical Quantities

Units And Measurements Problems In KSEEB Physics Chapter 2

Question 1. In the expression, P = El² m^-5 G^-2 the quantities E, l, m in and G denote energy, angular momentum, mass, and gravitational constant respectively. Show that P is a dimensionless, quantity.
Solution:

Here, P = El² m^-5 G^-2

Here, I = energy, l = angular momentum

m = mass, G = gravitational constant

= \(\left[M \mathrm{~L}^2 \mathrm{~T}^{-2}\right]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]^2[\mathrm{M}]^{-5}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-2}\)

= \(M^{1+2-5+2} \mathrm{~L}^{2+1-6} \mathrm{~T}^{-2-2+4}\)

P = \(\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]\)

Hence, P is a dimensionless quantity.

Question 2. If the velocity of light c, Planck’s constant, h, and the gravitational constant G are taken as fundamental quantities; then express mass, length, and time in terms of dimensions of these quantities.
Solution:

Here, \(c=\left[L \cdot T^{-1}\right] ; h=\left[M L^2 \mathrm{~T}^{-1}\right]\);

G = \(\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]\)

∴ E = \(\mathrm{h} v, \mathrm{~h}=\frac{\mathrm{E}}{v} ; \mathrm{G}=\left(\frac{\mathrm{Fd}^2}{\mathrm{~m}_1 \mathrm{~m}_2}\right)\)

Let m = \(c^x h^y G^z \rightarrow(1)\)

⇒ \(\left[M^1 L^0 T^0\right]=\left(L^{-1}\right)^x\left(M L^2 T^{-1}\right)^y\left(M^{-1} L^3 T^{-2}\right)^2\)

⇒ \(\left[M^1 L^0 T^0\right]=M^{y-z} L^{x+2 y+3 z} T^{-x-y-2 z}\)

Applying the principle of homogeneity of dimensions, we get

y – z = 1 → (2); x + 2y. + 3z = 0 → (3);

– x – y – 2z = 0 → (4)

Adding equation (2), equation (3) and equation (4),

2y = 1

⇒ y = \(\frac{1}{2}\)

∴ From eq. (2) z = y – 1 = \(\frac{1}{2}\) – 1 = \(\frac{-1}{2}\)

From eq. (4) x = – y – 2z = \(\frac{-1}{2}\) + 1 = \(\frac{1}{2}\)

Substituting the values of x, y, and z in eq. (1), we get

m = \(\mathrm{c}^{1 / 2} \mathrm{~h}^{1 / 2} \mathrm{G}^{-\sqrt{1 / 2}} \Rightarrow \mathrm{m}=\sqrt{\frac{\mathrm{ch}}{\mathrm{G}}}\)

Proceeding as above we can show that

L = \(\sqrt{\frac{\mathrm{hG}}{c^3}}\) and \(\mathrm{T}=\sqrt{\frac{\mathrm{hG}}{\mathrm{c}^5}}\)

Question 3. An artificial satellite revolves around a planet of mass M and radius R, in a circular orbit of radius r. Using dimensional analysis shows that the period of the satellite.

T = \(=\frac{k}{R} \sqrt{\frac{r^3}{g}}\)

where k is a dimensionless constant and g is the acceleration due to gravity.

Solution:

Given that T² ∝ r³ or T ∝ r^3/2

Also, T is a function of g and R

Let T ∝ r^3/2 g^a R^b where a, b are the dimensions of g and R.

(or) T = k r^3/2 g^a R^b → (1)

where k is a dimensionless constant of proportionality

From equation (1) \(\left[M^0 L^0 T^1\right]=L^{3 / 2}\left(L T^{-2}\right)^a(L)^b=M^0 L^{a+b+\frac{3}{2}} T^{-2 a}\)

Applying the principle of homogeneity of dimensions, we get

a + b + \(\frac{3}{2}=0\)…..(2)

∴ -2 a=1

⇒ a = \(\frac{-1}{2}\)

From eq (1), \(\frac{-1}{2}+b+\frac{3}{2}=0 \Rightarrow b=-1\)

Substituting the values of ‘a’ and ‘b’ in eq. (1), we get

T = \(k r^{3 / 2} \cdot g^{-1 / 2} R^{-1}\)

T = \(\frac{k}{R} \sqrt{\frac{r^3}{g}}\)

This is the required relation.

Units And Measurements KSEEB Class 11 Physics

Question 4. State the number of significant figures in the following

1. 6729
2. 0.024
3. 0.08240
4. 6.032
5. 4.57 x 10⁸

Solution:

1. In 6729 all are significant figures.
   • A number of significant figures are Four.
2. In 0.024 the zeroes to the left of 1st non-zero digit of a number less than one are not significant.
   • A number of significant figures Two.
3. 0.08240 – Significant figures Four.
   • In 6.032 the zero between two non-zero digits is significant.
   • So, a number of significant figures in 6.023 are 4.
4. 4.57 x 10⁸ Significant figures Three.

[In the representation of powers of Ten our rule is only significant figures must be given].

Question 5. One stick has a length of 12.132 cm and another has a length of 12.4 cm. If the two sticks are placed end what is the total length? If the two sticks are placed side by side, what Is the difference in their lengths?
Solution:

1. When placed end to end total length is

l = l₁ + l₂

l₁ = 12.132 cm and l₂ = 12.4 cm.

∴ l₁ + l₂ = 12.132 + 12.4 = 24.532 cm.

In addition, the final answer must have the least number of significant numbers in that addition, i.e., one after the decimal point.

So our answer is 24.5 cm.

2. For difference use l₁ – l₂

i.e., 12.4- 12.132 = 0.268 cm.

In subtraction final answer must be adjusted to the least number of significance figures in that operation.

Here least number is one digit after I decimal. By applying the round-off procedure] our answer is 0.3 cm.

Question 6. Each side of a cube is measured to be 7.203 m. What is

1. The total surface area and
2. The volume of the cube, to appropriate significant figures?

Solution:

Side of the cube, a = 7.203 m.

So a number of significant figures is Four.

1. Surface area of cube = 6a²

= 6x 7.203x 7.203 = 311.2991

But our final answer must be rounded to the least number of significant figures is four digits.

So the surface area of the cube = 311.3 m³

2. Volume of cube, V = a³ = (7.203)³ = 373.1471

But the answer must be limited to Four! significant figures.

∴ The volume of the sphere, V = 373.1 m³.

KSEEB Class 11 Physics solutions for Chapter 2 Units and Measurements

Question 7. The measured mass and volume of a body are 2.42 g and 4.7 cm³ respectively with possible errors 0.01 g and 0.1 cm³. Find the maximum error in density.
Solution:

Mass, m = 2.42 g ; Error, Δm = 0.01 g. Volume, V = 4.7 cm³, Error, ΔV = 0.1 cc.

% error in mass = \(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100=\frac{0.01}{2.42} \times 100\)

= \(\frac{1}{2.42}\)

% error in volume = \(\frac{0.1}{4.7} \times 100\) = \(\frac{10}{4.7}\)

Density = \(\frac{\text { Mass }}{\text { Volume }}=\frac{M}{V}\)

Maximum % error in density = % error in mass + % error in volume

∴ Maximum percentage error in density = \(\frac{1}{2.42}+\frac{10}{4.7}=0.413+2.127=2.54 \%\)

Question 8. The error in the measurement of the radius of a sphere is 1%. What is the error in the measurement of volume?
Solution:

Percentages Errors in radius = 1% = \(\frac{\Delta r}{r} \times 100\)

The volume of sphere V ∝ r³

⇒ ΔV = 3r²Δr

⇒ \(\frac{\Delta V}{V}=3 \frac{r^2 \cdot \Delta r}{r}\)

Percentage error in volume \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=3\left(\frac{\Delta \mathrm{r}}{\mathrm{r}} \times 100\right)=3 \times 1=3 \%\)

Question 9. The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities?
Solution:

Percentage change in mass = \(\frac{\Delta m}{m} \times 100\) = 2%

Percentages chage in speed = \(\frac{\Delta v}{v} \times 100\) = 3%

But K.E = \(\frac{1}{2}\) = mv

Percentages change in K.E = 1(% change mass) + 2(% change in velocity)

∴ Percentage change in K.E

= \(1\left(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100\right)+2\left(\frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100\right)\)

= 1 x 2 + 2 x 3 = 8%

Question 10. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). If the size of the hydrogen molecule is about 1Å, what is the ratio of molar volume to the atomic volume of a mole of hydrogen?
Solution:

Size of Hydrogen atom = 1Å =10^-10 m = 10^-8 cm

V₁ = Atomic volume = number of atoms x volume of atom.

One mole gas contains ‘n’ molecules.

Avogadro Number, n = 6.022 x 10²³

∴ \(V_1=\frac{4}{3} \pi r^3 \times n\)

= \(\frac{4}{3} \times \frac{22}{7} \times\left[10^{-8}\right]^3 \times 6.022 \times 10^{23}\)

= \(25.23 \times 10^{-1}\)or \(2.523 \mathrm{cc} \text {. }\)

∴ \(\mathrm{V}_2\) = Molar volume of 1 mol gas = 22.4 lit

= \(2.24 \times 10^4 \mathrm{c} . \mathrm{c}\)

∴ 1 lit =1000 c.c.

∴ The ratio of molar volume to atomic volume

= \(\mathrm{V}_2: \mathrm{V}_1\)

= \(2.24 \times 10^4: 2.523=10^4 \mathrm{~m}\)

• Chapter 1 Physical World Solutions
• Chapter 2 Units and Measurements
• Chapter 3 Motion in a Straight Line
• Chapter 4 Motion in a Plane
• Chapter 5 Laws of Motion
• Chapter 6 Work, Energy and Power
• Chapter 7 System of Particles and Rotational Motion

• Chapter 8 Gravitation
• Chapter 9 Mechanical Properties of Solids
• Chapter 10 Mechanical Properties of Fluids
• Chapter 11 Thermal Properties of Matter
• Chapter 12 Thermodynamics
• Chapter 13 Kinetic Theory
• Chapter 14 Oscillations
• Chapter 15 Waves