These Physics topics combine conceptual understanding with numerical problem-solving. Elasticity requires you to apply the relationships between force, deformation, and material properties, while KTG connects molecular motion with the macroscopic behaviour of gases. Thermodynamics involves tracking heat, work, and internal energy across different processes, so understanding sign conventions and interpreting PV graphs is essential.

For focused revision, Physics Wallah’s Thermodynamics, KTG, Heat, and Elasticity notes help you revisit the key formulas, standard processes, molecular speed relations, and elasticity applications covered in these topics. Use them to strengthen your understanding of the concepts and practise applying the relevant equations to numerical questions.

Elasticity: Stress, Strain, and Young’s Modulus

When an external force deforms a body, intermolecular forces develop within it and tend to restore its original shape and size. The ability of a body to regain its original dimensions after the deforming force is removed is called elasticity.

The force that acts to restore the original configuration is called the restoring force. At equilibrium, the restoring force balances the applied force.

Stress

Stress is the restoring force developed per unit area.

Stress = F/A

Here, F is the applied force and A is the cross-sectional area.

When the force acts perpendicular to the cross-section and causes a change in length, the stress is called normal stress or longitudinal stress.

Strain

Longitudinal strain is the ratio of the change in length to the original length.

Longitudinal strain = Change in length / Original length

Strain is dimensionless because it is the ratio of two quantities with the same unit.

Hooke’s Law and Young’s Modulus

Within the elastic limit, stress is directly proportional to strain. This relationship is known as Hooke’s law.

Young’s modulus measures a material’s resistance to longitudinal deformation.

Young’s modulus (Y) = Stress / Strain

For a wire of original length L and cross-sectional area A:

F/A = Y(ΔL/L)

Rearranging gives:

ΔL = FL/AY

Here, ΔL represents the change in length.

The extension of a wire increases with the applied force and original length. It decreases as the cross-sectional area or Young’s modulus increases.

Important comparison: Steel is more elastic than rubber in the physics sense because steel has a higher Young’s modulus.

Applications of Elasticity

Extension of a Suspended Wire

For a mass m suspended from a massless wire, the tension is:

T = mg

The extension is therefore:

ΔL = mgL/AY

This relationship is useful for numerical questions involving suspended masses and wire deformation.

Two Wires Connected in Series

When two wires are connected in series and carry the same tension, the extension of each wire depends on its length, area, and Young’s modulus.

For two wires:

ΔL₁/ΔL₂ = (L₁/L₂)(A₂/A₁)(Y₂/Y₁)

Use this relationship when comparing the extensions of two wire sections under the same tension.

Elastic Rods and Equivalent Springs

An elastic rod can be represented by an equivalent spring constant:

k = YA/L

For two springs connected in series, the equivalent spring constant is:

kₑq = k₁k₂/(k₁ + k₂)

When solving pulley-system questions, first determine the tension in each section. Then calculate the extension of each section using:

Extension of section i = TᵢLᵢ/(AᵢYᵢ)

Kinetic Theory of Gases

Kinetic Theory of Gases explains the macroscopic behaviour of gases in terms of the motion of their molecules.

The ideal-gas model assumes that:

• Molecules move randomly in all directions.

• The molecular volume is negligible compared with the volume occupied by the gas.

• Intermolecular forces are negligible except during collisions.

• Collisions between molecules and with the container walls are perfectly elastic.

• The time spent during collisions is negligible compared with the time between collisions.

An ideal gas is approximated most closely under conditions of relatively low pressure and high temperature. In the ideal-gas model, intermolecular potential energy is neglected.

Molecular Speeds

Gas molecules move at different speeds, so three measures are commonly used.

Average speed:

Average speed = √(8RT/πM)

Most probable speed:

Most probable speed = √(2RT/M)

Root mean square (RMS) speed:

RMS speed = √(3RT/M)

Here, R is the universal gas constant, T is the absolute temperature in kelvin, and M is the molar mass in kg/mol.

The three speeds follow this order:

RMS speed > Average speed > Most probable speed

The average velocity of gas molecules in a gas at equilibrium is zero because molecular velocities in different directions cancel out. Average speed is not zero because speed is a scalar quantity.

Effect of Temperature on Molecular Speed

All three characteristic molecular speeds are proportional to the square root of absolute temperature:

Speed ∝ √T

If the absolute temperature increases fourfold, the characteristic molecular speeds double.

Always convert temperature to kelvin before using these equations.

Thermodynamics: Heat, Work, and Internal Energy

The first law of thermodynamics expresses the conservation of energy in a thermodynamic system.

Using the convention that work done by the gas is positive:

Q = ΔU + W

Here:

• Q is the heat supplied to the gas.

• ΔU is the change in internal energy.

• W is the work done by the gas.

Heat supplied to the gas is positive, while heat released by the gas is negative. Work done during expansion is positive, and work done during compression is negative.

For an ideal gas, internal energy depends only on temperature. The change in internal energy is:

ΔU = nF RΔT/2

Here, n is the number of moles, F is the number of active degrees of freedom, R is the universal gas constant, and ΔT is the change in temperature.

Important Questions

1. A solid sphere of radius R is made of a soft material with bulk modulus K. It is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid, covering the entire cross-section of the cylindrical container. When a mass m is placed on the piston to compress the liquid, the fractional change in the radius of the sphere, R/R, is:

Answer:

δR/R = −mg/(3AK) 

Explanation: The applied mass increases the pressure in the liquid by

P=mgA

The bulk modulus is defined as

K=-PV/V

For a sphere, V=43R3. For a small change in radius,

VV=3RR

Therefore,

δR/R = −mg/(3AK) 

The negative sign indicates that the radius decreases due to compression.

2. The length of a string is 1.4 m when the tension is 5 N and 1.56 m when the tension is 7 N. Find the original length of the string.

Answer: 1.0m

Explanation: According to Hooke’s law, the extension of a string is proportional to the applied tension, provided the elastic limit is not exceeded.

Let L0 be the original length and k be the extension per unit tension.

1.4=L0+5k

1.56=L0+7k

Subtracting the first equation from the second:

0.16=2k

k=0.08m/N

Substituting into the first equation:

L0=1.4-5(0.08)

L₀ = 1.0 m 

3. The Young’s modulus of the material of a wire is Y, its length is L, and its cross-sectional area is A. If the length of the wire is doubled and its cross-sectional area is halved, then the Young’s modulus is:

Answer: Y

Explanation: Young’s modulus is a property of the material and is defined as the ratio of longitudinal stress to longitudinal strain. It does not depend on the dimensions of the wire, provided the material and physical conditions remain unchanged.

4. One end of a metal wire is fixed to the ceiling, and a load of 2 kg hangs from its other end. A similar wire is attached to the bottom of this load, and another load of 1 kg hangs from the lower wire. Find the ratio of the longitudinal strain of the upper wire to that of the lower wire.

Answer: 3:1

Explanation: Longitudinal strain is given by

Strain=StressY=FAY

Since both wires are similar, they have the same cross-sectional area and Young’s modulus. Therefore, the ratio of their strains equals the ratio of the tensions.

• Tension in the upper wire: T1=(2+1)g=3g

• Tension in the lower wire: T2=1g=g

Thus,

12=T1T2=3gg

ε₁ : ε₂ = 3 : 1 

5. Find the RMS velocity of an oxygen molecule at \(27^\circ\text{C}\).

Answer: Approximately 483m/s

Explanation: The RMS velocity of a gas molecule is given by

vrms=3RTM

Where:

• R=8.314Jmol-1K-1

• T=27+273=300K

• M=3210-3kg/mol for oxygen gas

Substituting the values:

vrms=38.3143000.032

vᵣₘₛ ≈ 483 m/s 

6. Diatomic gases are compressed adiabatically to 1/32 of their initial volume. If the initial temperature is T0, find the final temperature. (JEE Advanced 2010) 

Answer: 4T0

Explanation: For an adiabatic process involving an ideal gas,

TV-1=constant

For a diatomic gas at ordinary temperatures,

=75

Therefore,

T0V0-1=TfVf-1

Given Vf=V0/32,

TfT0=V0Vf-1

TfT0=322/5

Since 32=25,

322/5=22=4

T_f = 4T₀ 

7. A gas expands isothermally along AB and does 700 J of work.

(a) How much heat does the gas exchange along AB?

(b) If the gas then expands adiabatically along BC and does 400 J of work, what happens to its temperature?

(c) When the gas returns to A along CA, it releases 100 J of heat to the surroundings. How much work is done on the gas along this path?

Answer:

• (a) QAB=+700J

• (b) The temperature decreases.

• (c) 500J of work is done on the gas.

Explanation: According to the first law of thermodynamics,

U=Q-W

where Q is the heat supplied to the gas and W is the work done by the gas.

(a) Isothermal expansion along AB

For an ideal gas, internal energy depends only on temperature. Since the temperature remains constant,

U=0

Therefore,

Q=W=700J

The gas absorbs 700 J of heat.

(b) Adiabatic expansion along BC

In an adiabatic process, no heat is exchanged.

Q=0,U=-W=-400J

The internal energy decreases, so the temperature of the ideal gas decreases.

(c) Return along CA

Since the gas releases 100 J of heat,

Q=-100J

Over a complete cycle, the net change in internal energy is zero. Hence,

UCA=-(0-400)=400J

Using the first law:

400=-100-W

W=-500J

The negative sign indicates that work is done on the gas.

W_on = 500 J 

8. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case?

Answer: Approximately 16.8J of work is done by the system.

Explanation: The change in internal energy depends only on the initial and final states. Therefore, it is the same for both processes.

For the adiabatic process,

Q=0

Since 22.3 J of work is done on the system,

U=22.3J

For the second process, convert heat into joules:

Q=9.354.184

Q39.13J

Using the first law,

U=Q-W

22.3=39.13-W

W=39.13-22.3

W ≈ 16.8 J 

Thermodynamics, KTG, Heat, and Elasticity become easier to revise when you understand the physical meaning behind each formula and know when to apply it. Focus on wire deformation, elastic moduli, molecular speeds, heat capacities, thermodynamic processes, and the interpretation of PV graphs. Physics Wallah’s Thermodynamics, KTG, Heat, and Elasticity one-shot notes can help you review these concepts and practise applying them to numerical questions.

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