A Carnot engine operates between temperatures 400 K and 300 K. Its maximum efficiency is:

The correct answer is A: 25%. Carnot efficiency η = 1 - (T_cold/T_hot) = 1 - (300/400) = 1 - 0.75 = 0.25 = 25%

A substance has ΔH = -150 kJ/mol and ΔS = -100 J/(mol·K). At what temperature will ΔG = 0?

The correct answer is A: 1500 K. At equilibrium: ΔG = 0, so ΔH = TΔS. T = ΔH/ΔS = (-150000)/(-100) = 1500 K

The coefficient of performance (COP) of a refrigerator is 5. The work input required to remove 500 J of heat is:

The correct answer is A: 100 J. COP = Q_removed/W_input, so W = Q/COP = 500/5 = 100 J

Which of the following represents a scalar quantity?

The correct answer is C: Mass. Mass is a scalar quantity as it has magnitude only, not direction. Displacement, velocity, and acceleration are vectors.

An object moving at 15 m/s decelerates uniformly and stops after traveling 50 m. What is the deceleration?

The correct answer is C: 2.25 m/s². Using v² = u² - 2as, 0 = 225 - 2a(50), therefore a = 225/100 = 2.25 m/s²

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NEET Physics

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

70 Q 2 Topics Take Mock Test

Difficulty: All Easy Medium Hard 21–30 of 70

Topics in NEET Physics

All Mechanics & Laws of Motion 100 Thermodynamics 100

Q.21 Easy Thermodynamics

The maximum possible efficiency of a heat engine operating between 400 K and 500 K is:

A 11%

B 20%

C 25%

D 80%

Correct Answer: B. 20%

EXPLANATION

Maximum efficiency is Carnot efficiency: η_max = 1 - T_c/T_h = 1 - (400/500) = 1 - 0.8 = 0.2 = 20%.

Test

Q.22 Easy Thermodynamics

A heat engine absorbs 500 J of heat and does 200 J of work. The heat rejected is:

A 700 J

B 300 J

C 200 J

D 500 J

Correct Answer: B. 300 J

EXPLANATION

By energy conservation: Q_in = W + Q_out, so Q_out = Q_in - W = 500 - 200 = 300 J.

Test

Q.23 Easy Thermodynamics

For an isothermal process of an ideal gas, the relationship between pressure and volume is:

A PV = constant

B P/V = constant

C P + V = constant

D PV² = constant

Correct Answer: A. PV = constant

EXPLANATION

In isothermal process, T is constant. From ideal gas law PV = nRT, if T is constant, then PV = constant (Boyle's Law).

Test

Q.24 Easy Thermodynamics

Which process has the property that pressure remains constant?

A Isochoric

B Isobaric

C Isothermal

D Adiabatic

Correct Answer: B. Isobaric

EXPLANATION

Isobaric process is defined as a process at constant pressure. Prefix 'iso' means constant, and 'baric' refers to pressure.

Test

Q.25 Easy Thermodynamics

A Carnot engine operates between 300 K and 600 K. Its efficiency is:

A 25%

B 50%

C 75%

D 100%

Correct Answer: B. 50%

EXPLANATION

Carnot efficiency η = 1 - (T_cold/T_hot) = 1 - (300/600) = 1 - 0.5 = 0.5 = 50%.

Test

Q.26 Easy Thermodynamics

For a cyclic process, which of the following is always true?

A ΔU = 0 and Q = W

B ΔU ≠ 0 but Q = 0

C W = 0 but Q ≠ 0

D Both Q and W are zero

Correct Answer: A. ΔU = 0 and Q = W

EXPLANATION

In a cyclic process, the system returns to its initial state, so ΔU = 0. By first law: Q = ΔU + W = W.

Test

Q.27 Easy Thermodynamics

A gas undergoes an adiabatic compression. Which statement is correct?

A Temperature increases and internal energy decreases

B Temperature increases and internal energy increases

C Temperature decreases and internal energy increases

D Temperature and internal energy both decrease

Correct Answer: B. Temperature increases and internal energy increases

EXPLANATION

In adiabatic compression, Q=0. Work is done ON the gas (W<0), so ΔU = Q - W = -W > 0. Internal energy increases, causing temperature to rise.

Test

Q.28 Easy Thermodynamics

A gas expands from volume V to 2V at constant pressure P. The work done by the gas is:

A PV

B 2PV

C P(2V - V) = PV

D P/2V

Correct Answer: C. P(2V - V) = PV

EXPLANATION

Work in isobaric process: W = P(V_f - V_i) = P(2V - V) = PV

Test

Q.29 Easy Thermodynamics

In an isochoric process, the work done by the gas is:

A Maximum

B Minimum

C Zero

D Equal to heat absorbed

Correct Answer: C. Zero

EXPLANATION

In isochoric (constant volume) process, W = ∫PdV = 0 since dV = 0

Test

Q.30 Easy Thermodynamics

For an ideal gas undergoing isobaric process, the relation between volume and temperature is:

A V ∝ T

B V ∝ 1/T

C V ∝ T²

D V ∝ √T

Correct Answer: A. V ∝ T

EXPLANATION

Charles's Law: At constant pressure, V/T = constant, so V ∝ T

Test

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