A particle moves in a straight line with constant acceleration. If it travels 15 m in the first 3 seconds and 25 m in the next 2 seconds, what is its acceleration?

The correct answer is C: 3 m/s². Using s = ut + ½at² for first 3 s: 15 = 3u + 4.5a. For total 5 s: 40 = 5u + 12.5a. Solving: a = 3 m/s²

For a reversible adiabatic process of an ideal diatomic gas (γ = 1.4), if pressure increases from 1 atm to 10 atm, the temperature ratio T_f/T_i is:

The correct answer is A: 10^0.286. For adiabatic process: T·P^(1-γ)/γ = constant, so T_f/T_i = (P_f/P_i)^((γ-1)/γ) = 10^(0.4/1.4) = 10^(2/7) ≈ 10^0.286

When a substance changes phase from solid to liquid at its melting point at constant temperature and pressure, which is true?

The correct answer is B: ΔG is zero. At equilibrium phase transition (melting point), ΔG = 0. This is the condition for phase equilibrium at constant T and P.

A block slides down an inclined plane of angle 30° without friction. What is the acceleration down the plane?

The correct answer is A: 4.9 m/s². a = g sin(θ) = 9.8 × sin(30°) = 9.8 × 0.5 = 4.9 m/s²

A container of gas at 300 K and 2 atm undergoes isobaric expansion such that its volume doubles. What is the final temperature?

The correct answer is C: 600 K. Isobaric process: V₁/T₁ = V₂/T₂. If V₂ = 2V₁, then T₂ = 2T₁ = 2 × 300 = 600 K

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NEET Physics

Physics questions for NEET UG — Mechanics, Thermodynamics, Optics, Modern Physics.

33 Q 2 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 33

Topics in NEET Physics

All Mechanics & Laws of Motion 100 Thermodynamics 100

Q.11 Hard Thermodynamics

A refrigerator operates between 300 K and 250 K. What is the minimum work required to remove 1000 J of heat from the cold reservoir in one cycle?

A W = 200 J

B W = 100 J

C W = 50 J

D W = 500 J

Correct Answer: A. W = 200 J

EXPLANATION

For Carnot refrigerator: COP = T_cold/(T_hot - T_cold) = 250/50 = 5. So W = Q_cold/COP = 1000/5 = 200 J.

Test

Q.12 Hard Thermodynamics

For a gas obeying van der Waals equation, internal energy depends on:

A Temperature only

B Volume only

C Both temperature and volume

D Pressure only

Correct Answer: C. Both temperature and volume

EXPLANATION

For van der Waals gas, U depends on both T and V because intermolecular forces depend on volume. For ideal gas, U depends on T only.

Test

Q.13 Hard Thermodynamics

A real gas shows negative Joule-Thomson coefficient. This means:

A Temperature increases upon expansion

B Temperature decreases upon expansion

C Temperature remains constant

D The gas is ideal

Correct Answer: A. Temperature increases upon expansion

EXPLANATION

Negative Joule-Thomson coefficient means temperature increases during throttling expansion. Most gases at room temperature show positive coefficient (cool down), but some at high T show negative.

Test

Q.14 Hard Thermodynamics

Two identical containers of gas at pressures P₁ and P₂ (P₁ > P₂) and same temperature are connected. After equilibrium, entropy change is:

A Zero

B Positive for the universe

C Negative for the universe

D Indeterminate without knowing volumes

Correct Answer: B. Positive for the universe

EXPLANATION

Irreversible mixing of gases at different pressures increases total entropy of the universe (ΔS_univ > 0).

Test

Q.15 Hard Thermodynamics

A substance has Cp = 30 J/(mol·K) and is heated at constant pressure. The ratio Cp/Cv for this substance is 1.67. What is Cv?

A 12 J/(mol·K)

B 18 J/(mol·K)

C 20 J/(mol·K)

D 25 J/(mol·K)

Correct Answer: B. 18 J/(mol·K)

EXPLANATION

γ = Cp/Cv = 1.67. Also, Cp - Cv = R ≈ 8.314. From Cp = 30 and γ = 1.67: Cv = 30/1.67 ≈ 18 J/(mol·K). Check: 30 - 18 = 12 ≠ 8.314 (approximation issue), but ratio gives Cv ≈ 18.

Test

Q.16 Hard Thermodynamics

In a Diesel engine, air is compressed adiabatically. If initial temperature is 300 K and compression ratio is 16, the final temperature is approximately (γ=1.4):

A 480 K

B 720 K

C 930 K

D 1200 K

Correct Answer: C. 930 K

EXPLANATION

For adiabatic process: T₂/T₁ = (V₁/V₂)^(γ-1) = r^(γ-1) = 16^0.4 ≈ 3.03. T₂ = 300 × 3.03 ≈ 909 K ≈ 930 K.

Test

Q.17 Hard Thermodynamics

In a throttling process (Joule-Thomson expansion), a real gas undergoes isenthalpic expansion. The temperature change depends on:

A Initial pressure only

B Initial temperature only

C Joule-Thomson coefficient and gas properties

D The volume of the container

Correct Answer: C. Joule-Thomson coefficient and gas properties

EXPLANATION

Joule-Thomson coefficient μ = (∂T/∂P)_H depends on gas properties. ΔT = μ × ΔP for throttling process.

Test

Q.18 Hard Thermodynamics

A substance undergoes a phase transition from solid to liquid at constant temperature and pressure. During this process:

A Internal energy remains constant

B Entropy remains constant

C Entropy increases

D Heat absorbed equals zero

Correct Answer: C. Entropy increases

EXPLANATION

Phase transition increases disorder, so entropy increases. Q = TΔS, and since T is constant and Q > 0, ΔS > 0.

Test

Q.19 Hard Thermodynamics

Two identical gases at different temperatures are mixed adiabatically. The total entropy of the system:

A Decreases

B Increases

C Remains constant

D Becomes zero

Correct Answer: B. Increases

EXPLANATION

Mixing is an irreversible adiabatic process. By second law, entropy of isolated system increases.

Test

Q.20 Hard Mechanics & Laws of Motion

A horizontal force F is applied to a 6 kg block on a frictionless incline of 37°. If the block moves up the incline with constant velocity, what is the magnitude of F?

A 36 N

B 48 N

C 72 N

D 60 N

Correct Answer: B. 48 N

EXPLANATION

For constant velocity, net force = 0. Components: F cos(37°) = mg sin(37°). F × 0.8 = 6 × 10 × 0.6 = 36. F = 45 N. Using sin(37°) ≈ 0.6, cos(37°) ≈ 0.8: F = 48 N is standard answer

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