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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

= \((5 \times 10 \times 10)+\frac{1}{\not 2} \times 5 \times 20^{10} \times 20\)

= \(500+1000=1500 \mathrm{~J}\)