The EMF of a cell is 1.5V. When a current of 2A is drawn, the terminal voltage is 1.2V. The internal resistance is:

The correct answer is A: 0.15Ω. E = V + Ir → 1.5 = 1.2 + 2r → 0.3 = 2r → r = 0.15Ω.

An LED emits light because:

The correct answer is A: Holes and electrons recombine radiatively in direct bandgap material. In LEDs (forward biased p-n junctions in direct bandgap semiconductors like GaAs), electron-hole recombination releases energy as photons. Indirect bandgap materials (Si, Ge) produce mainly heat.

A conducting rod of length L moves with velocity v perpendicular to a magnetic field B. The induced EMF across the rod is:

The correct answer is A: BLv. Motional EMF = BLv (where L is perpendicular to both B and v)

A uniform rod of mass M and length L is pivoted at one end. The moment of inertia about the pivot is:

The correct answer is A: ML²/3. For a uniform rod about one end, I = ML²/3 (standard formula)

The conductivity of a semiconductor is given by σ = e(neμe + nhμh). If temperature increases from 300K to 400K, which factor primarily determines the change in conductivity?

The correct answer is A: Increase in carrier concentration dominates. Though mobility decreases with temperature (T^-3/2), the exponential increase in carrier concentration (proportional to exp(-Eg/2kT)) dominates, resulting in net increase in conductivity.

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JEE Physics
Thermodynamics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

50 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 31–40 of 50

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.31 Medium Thermodynamics

Two identical blocks of metal at different temperatures are brought into thermal contact in an isolated system. The process is:

A Reversible and entropy of universe increases

B Reversible and entropy of universe remains constant

C Irreversible and entropy of universe increases

D Irreversible and entropy of universe remains constant

Correct Answer: C. Irreversible and entropy of universe increases

EXPLANATION

Heat transfer between objects at different temperatures is irreversible. For an isolated system, ΔS_universe = ΔS_system > 0 (irreversible process), not equal to zero.

Test

Q.32 Medium Thermodynamics

One mole of an ideal monoatomic gas is heated at constant volume from 300 K to 600 K. The change in internal energy is (R = 8.314 J/mol·K):

A 12,471 J

B 24,942 J

C 37,413 J

D 6,235.5 J

Correct Answer: A. 12,471 J

EXPLANATION

For monoatomic ideal gas, Cv = (3/2)R. ΔU = nCvΔT = 1 × (3/2) × 8.314 × (600 − 300) = (3/2) × 8.314 × 300 = 3,741 × 3.33 ≈ 12,471 J

Test

Q.33 Medium Thermodynamics

A refrigerator operates on a Carnot cycle between -10°C and 30°C. The coefficient of performance (COP) is approximately:

A 4.5

B 6.5

C 8.5

D 10.5

Correct Answer: B. 6.5

EXPLANATION

COP = Tc/(Th - Tc) = 263/(303 - 263) = 263/40 ≈ 6.575 ≈ 6.5

Test

Q.34 Medium Thermodynamics

The molar heat capacity at constant pressure for a diatomic ideal gas is:

A (3/2)R

B (5/2)R

C (7/2)R

D (9/2)R

Correct Answer: C. (7/2)R

EXPLANATION

For diatomic gas: Cv = (5/2)R. Using Cp = Cv + R, we get Cp = (5/2)R + R = (7/2)R

Test

Q.35 Medium Thermodynamics

A cylinder with a movable piston contains 2 moles of ideal gas at 300 K and 1 atm. When heated at constant pressure to 600 K, what is the work done by the gas? (R = 8.314 J/mol·K)

A 2492 J

B 4984 J

C 9968 J

D 1246 J

Correct Answer: B. 4984 J

EXPLANATION

For isobaric process: W = nRΔT = 2 × 8.314 × (600 - 300) = 2 × 8.314 × 300 = 4988.4 ≈ 4984 J

Test

Q.36 Medium Thermodynamics

Which process would result in the maximum work output from an ideal gas expansion between two fixed pressures?

A Isothermal process

B Isobaric process

C Isochoric process

D Adiabatic process

Correct Answer: A. Isothermal process

EXPLANATION

For expansion between fixed initial and final states, isothermal process yields maximum work because the gas maintains maximum pressure throughout the expansion compared to adiabatic or other processes.

Test

Q.37 Medium Thermodynamics

The entropy change for a reversible isothermal process is given by:

A ΔS = Q/T

B ΔS = nR ln(Vf/Vi)

C Both A and B

D ΔS = nCv ln(Tf/Ti)

Correct Answer: C. Both A and B

EXPLANATION

For reversible isothermal process: ΔS = Q/T = nR ln(Vf/Vi) since Q = nRT ln(Vf/Vi) for ideal gas isothermal expansion.

Test

Q.38 Medium Thermodynamics

A heat engine absorbs 1000 J of heat and rejects 600 J to the cold reservoir in one cycle. What is its efficiency?

A 40%

B 60%

C 37.5%

D 62.5%

Correct Answer: A. 40%

EXPLANATION

Efficiency η = (Qh - Qc)/Qh = (1000 - 600)/1000 = 400/1000 = 0.40 = 40%

Test

Q.39 Medium Thermodynamics

Two samples of the same ideal gas at the same temperature have volumes V and 2V respectively. The ratio of their internal energies is:

A 1:2

B 1:1

C 2:1

D Cannot be determined

Correct Answer: D. Cannot be determined

EXPLANATION

Internal energy U = nCvT. Without knowing the number of moles in each sample, the ratio cannot be determined. Same temperature doesn't mean same internal energy.

Test

Q.40 Medium Thermodynamics

For a diatomic ideal gas undergoing an isothermal process, which quantity remains constant?

A Pressure

B Volume

C Internal energy

D Density

Correct Answer: C. Internal energy

EXPLANATION

In an isothermal process, temperature is constant. For an ideal gas, internal energy depends only on temperature, so ΔU = 0. Pressure and volume change according to PV = constant.

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