Practice <strong>NEET UG Physics</strong> MCQ questions covering Mechanics, Thermodynamics, Waves & Optics, Modern Physics, Electricity & Magnetism, and Electromagnetic Induction. Aligned with NCERT Class 11 & 12 Physics syllabus.
The coefficient of performance (COP) of a refrigerator is 5. The work input required to remove 500 J of heat is:
Answer: A
COP = Q_removed/W_input, so W = Q/COP = 5500 = 100 J
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Q.6Medium
At absolute zero, the entropy of a perfect crystal is:
Answer: B
According to third law of thermodynamics, entropy of a perfect crystal at 0 K is zero.
Q.7Medium
A reversible engine and an irreversible engine operate between the same two temperatures. Which has greater efficiency?
Answer: A
Carnot (reversible) engine has maximum efficiency between any two temperatures. All real irreversible engines are less efficient.
Q.8Medium
The heat capacity at constant pressure (Cp) for an ideal gas is greater than at constant volume (Cv) because:
Answer: B
At constant P, heat goes into both increasing internal energy and doing work: Q_p = ΔU + W. At constant V, heat only increases internal energy: Q_v = ΔU. Thus Cp > Cv by R.
Q.9Medium
For one mole of ideal gas, Cp - Cv equals:
Answer: B
Mayer's relation: Cp - Cv = R for one mole of ideal gas (in terms of molar heat capacities)
Q.10Medium
The entropy change of the universe in an irreversible process is:
Answer: C
By the second law of thermodynamics, entropy of an isolated system increases for irreversible processes (ΔS_universe > 0).
Q.11Medium
For a reversible adiabatic process of an ideal gas, which relation holds?
Answer: A
For adiabatic process: PV^γ = constant. Using ideal gas law PV = nRT, we derive TV^(γ-1) = constant.
Q.12Medium
A refrigerator with COP = 4 requires 100 J of work per cycle. Heat removed from the cold reservoir is:
Answer: C
COP = Q_c/W, where Q_c is heat removed from cold reservoir. Q_c = COP × W = 4 × 100 = 400 J.
Q.13Medium
When ice melts at 0°C (273 K) at atmospheric pressure, the entropy change is related to:
Answer: B
For phase transition at constant T and P: ΔS = Q_rev/T = L_f/T, where L_f is latent heat of fusion.
Q.14Medium
A gas expands against a constant external pressure of 1 atm from 1 L to 5 L. The work done by the gas is:
Answer: B
W = P_ext × ΔV = 1 atm × (5-1) L = 4 L·atm = 4 × 101.325 = 405 J (positive, work done by gas).
Q.15Medium
Which of the following is NOT a state function?
Answer: C
Heat (Q) and work (W) are path functions, not state functions. They depend on the process, not just initial and final states.
Q.16Medium
The first law of thermodynamics can be written as dU = δQ - δW. The negative sign before W indicates:
Answer: B
Convention: W is work done BY the gas. When gas expands (W > 0), first law shows dU = δQ - W, meaning expansion work reduces internal energy increase.
Q.17Medium
For a Carnot engine operating between temperatures T₁ (hot) and T₂ (cold), the maximum efficiency is:
Answer: A
Maximum (Carnot) efficiency: η_max = 1 - T_cold/T_hot = 1 - T₂/T₁. This is the theoretical maximum for any heat engine.
Q.18Medium
A system absorbs 1000 J of heat and does 600 J of work on the surroundings. What is the change in internal energy?
Answer: A
ΔU = Q - W = 1000 - 600 = 400 J. When work is done BY the system, it's subtracted from heat absorbed.
Q.19Medium
What is the relationship between Cp and Cv for an ideal gas?
Answer: D
Both relationships are true: Cp - Cv = R (molar basis) and Cp/Cv = γ. For monoatomic gas, γ = 35; for diatomic, γ = 57.
Q.20Medium
In an adiabatic expansion of an ideal gas, the temperature decreases. This is because:
Answer: B
In adiabatic process, Q = 0. Since ΔU = -W and W > 0 (expansion), ΔU < 0, so temperature decreases. Internal energy decreases as work is done by the gas.
