Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
A gas undergoes an isothermal process at 300 K. If the initial volume is 2 L and final volume is 5 L, what is the work done by the gas? (Take R = 8.314 J/(mol·K), n = 1 mol)
A Carnot engine operates between temperature reservoirs of 500 K and 300 K. What is its maximum efficiency?
Answer: A
Carnot efficiency: η = 1 - (Tc/Th) = 1 - (500300) = 1 - 0.60 = 0.40 or 40%
Q.3Medium
In an adiabatic process with a diatomic ideal gas (γ = 1.4), if the initial temperature is 300 K and volume increases by factor of 2, what is the final temperature?
The first law states: ΔU = Q + W, where ΔU is change in internal energy, Q is heat absorbed, and W is work done on the system
Q.5Medium
A cyclic process consists of two isothermal and two adiabatic processes (Carnot cycle). What is true about the heat exchanges?
Answer: A
In Carnot cycle, isothermal processes involve heat exchange with reservoirs (Q₁ at hot reservoir, Q₂ at cold reservoir), while adiabatic processes have no heat exchange (Q=0)
Q.6Easy
For an ideal gas, internal energy depends on:
Answer: A
For an ideal gas, internal energy U depends only on temperature (U = nCᵥT), not on pressure or volume independently
Q.7Medium
In an isobaric process, 5 moles of a monoatomic ideal gas are heated from 300 K to 400 K. What is the heat supplied? (R = 8.314 J/(mol·K))
Three identical conducting rods are arranged in series between two heat reservoirs at 100°C and 0°C. At steady state, what is the temperature at the junction between the second and third rod?
Answer: A
In series arrangement with identical rods, temperature difference is equally distributed. ΔT_total = 100°C, so ΔT per rod = 3100 = 33.3°C. Second junction = 100 - 2(33.3) = 33.3°C
Q.14Easy
What is the Clausius statement of the second law of thermodynamics?
Answer: A
Clausius statement: Heat cannot spontaneously transfer from a colder body to a hotter body without external work being done on the system
Q.15Hard
A gas undergoes a cyclic process ABCA where AB is isothermal, BC is adiabatic, and CA is isochoric. If work is done on the gas in the cycle, what can be concluded?
Answer: A
If W_net < 0 (work done on gas), then from first law: ΔU_cycle = 0 = Q - W, so Q = W < 0, meaning net heat flows out
Q.16Medium
For a van der Waals gas, which statement is correct?
Answer: A
Van der Waals equation (P + a/V²)(V - b) = RT accounts for molecular volume (b term) and intermolecular attractive forces (a term)
Q.17Medium
In an expansion process, a gas does 500 J of work and absorbs 300 J of heat. What is the change in internal energy?
Answer: A
Using first law: ΔU = Q - W = 300 - 500 = -200 J (internal energy decreases)
Q.18Easy
For a diatomic ideal gas at room temperature, what is the ratio γ = Cₚ/Cᵥ?
Answer: A
For diatomic gas: Cᵥ = (25)R and Cₚ = (27)R. γ = Cₚ/Cᵥ = 57 = 1.40
Q.19Hard
Two bodies at temperatures T₁ = 400 K and T₂ = 300 K are brought into thermal contact. If entropy change of universe is 0.575 J/K and heat capacity of both bodies is 1000 J/K, what is the final equilibrium temperature? (Assume no heat loss to surroundings)
Answer: B
Heat lost by body 1: Q = C(T₁ - T_f) = 1000(400 - T_f). Heat gained by body 2: Q = 1000(T_f - 300). ΔS_univ = C ln(T_f/T₁) + C ln(T_f/T₂) = 1000[ln(T_f/400) + ln(T_f/300)] = 0.575. Solving: T_f = 350 K
Q.20Easy
A thermodynamic system undergoes a process where internal energy increases by 150 J while the system does 100 J of work on surroundings. What is the heat absorbed by the system?
Answer: A
By first law: ΔU = Q - W. Here ΔU = 150 J, W = 100 J (work done by system). So Q = ΔU + W = 150 + 100 = 250 J
