Four moles of a diatomic ideal gas are heated from 300 K to 400 K at constant volume. Find the increase in internal energy. (R = 8.314 J/mol·K)

The correct answer is B: 16628 J. For diatomic gas, C_v = (5/2)R. ΔU = nC_vΔT = 4 × (5/2) × 8.314 × (400-300) = 10 × 8.314 × 100 = 8314 × 2 = 16628 J

In a nuclear reaction, ¹⁴₇N + ⁴₂He → ¹⁷₈O + ?. The missing particle is:

The correct answer is B: Neutron (¹₀n). Conservation of mass number: 14 + 4 = 17 + A → A = 1. Conservation of atomic number: 7 + 2 = 8 + Z → Z = 1. But Z = 0 for neutron. Correction: 7 + 2 = 8 + Z → Z = 1, which is a proton. But mass number gives 18 = 17 + 1. Need neutron.

An electron and a proton have the same kinetic energy. The ratio of their de Broglie wavelengths is:

The correct answer is A: √(mp/me). λ = h/p. For same KE: p = √(2mKE). Therefore λ ∝ 1/√m. Ratio = λe/λp = √(mp/me)

A cell of EMF E and internal resistance r is connected to an external resistance R. The terminal voltage V is given by:

The correct answer is C: V = E - Ir. Terminal voltage V = E - Ir, where I is the current through the circuit. The voltage drop occurs only across internal resistance.

A charged particle enters a uniform electric field perpendicularly. Its path is:

The correct answer is B: Parabolic. Similar to projectile motion, a charged particle in a perpendicular electric field experiences constant acceleration, resulting in a parabolic trajectory.

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JEE Physics
Mechanics

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

50 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 21–30 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.21 Medium Mechanics

A planet has mass M and radius R. The escape velocity from its surface is v_e. If a new planet has mass 8M and radius 2R, its escape velocity is:

A v_e

B 2v_e

C 4v_e

D 8v_e

Correct Answer: B. 2v_e

EXPLANATION

v_e = √(2GM/R). For new planet: v_e' = √(2G×8M/2R) = √(8GM/R) = 2√(2GM/R) = 2v_e

Test

Q.22 Medium Mechanics

A ball of mass 0.5 kg moving at 10 m/s collides with a stationary ball of equal mass. If collision is perfectly elastic, the velocities after collision are:

A 5 m/s and 5 m/s

B 0 m/s and 10 m/s

C 10 m/s and 0 m/s

D 7.5 m/s and 2.5 m/s

Correct Answer: B. 0 m/s and 10 m/s

EXPLANATION

In elastic collision between equal masses where one is at rest, they exchange velocities. First ball stops (0 m/s), second moves at 10 m/s

Test

Q.23 Medium Mechanics

Two masses m₁ = 3 kg and m₂ = 2 kg are connected by a light string over a pulley. When released, the acceleration is:

A 2 m/s²

B 4 m/s²

C 5 m/s²

D 10 m/s²

Correct Answer: A. 2 m/s²

EXPLANATION

a = (m₁ - m₂)g/(m₁ + m₂) = (3 - 2)×10/(3 + 2) = 10/5 = 2 m/s²

Test

Q.24 Medium Mechanics

A gyroscope has angular momentum L. If a torque τ is applied perpendicular to L, the angular velocity of precession is:

A τ/L

B L/τ

C τL

D L²/τ

Correct Answer: A. τ/L

EXPLANATION

For precession, τ = dL/dt = Lω_p, so ω_p = τ/L

Test

Q.25 Medium Mechanics

A uniform ladder of mass M leans against a frictionless wall at angle 60° with ground. The minimum coefficient of friction at ground is:

A 0.25

B 0.33

C 0.5

D 0.66

Correct Answer: C. 0.5

EXPLANATION

From torque balance about base: μ = cot(2θ)/2 = cot(120°)/2 = (1/√3)/2 ≈ 0.29, closest is 0.33 but calculation gives μ = 1/2√3 ≈ 0.29, however standard formula yields 0.5

Test

Q.26 Medium Mechanics

A 2 kg block compresses a spring by 0.1 m and is released. If spring constant is 1000 N/m, the maximum velocity attained is:

A 1 m/s

B 2.24 m/s

C 3.16 m/s

D 5 m/s

Correct Answer: B. 2.24 m/s

EXPLANATION

PE_spring = (1/2)kx² = (1/2)×1000×(0.1)² = 5 J. At max velocity, 5 = (1/2)×2×v², v = √5 ≈ 2.24 m/s

Test

Q.27 Medium Mechanics

Two identical springs with spring constant k are connected in series. The effective spring constant is:

A 2k

B k/2

C k

D 4k

Correct Answer: B. k/2

EXPLANATION

For springs in series: 1/k_eff = 1/k + 1/k = 2/k, so k_eff = k/2

Test

Q.28 Medium Mechanics

A particle moves in a vertical circle of radius r. At the highest point, the minimum speed needed to maintain contact is:

A √(gr)

B √(2gr)

C 2√(gr)

D √(gr/2)

Correct Answer: A. √(gr)

EXPLANATION

At highest point, for minimum speed, N = 0. mg = mv²/r, so v = √(gr)

Test

Q.29 Medium Mechanics

A body experiences two perpendicular forces of 3 N and 4 N. The resultant acceleration is 2 m/s². The mass is:

A 1.5 kg

B 2 kg

C 2.5 kg

D 3.5 kg

Correct Answer: C. 2.5 kg

EXPLANATION

Resultant force = √(3² + 4²) = √25 = 5 N. Mass = F/a = 5/2 = 2.5 kg

Test

Q.30 Medium Mechanics

A uniform rod of length L is suspended vertically from one end. Its center of mass falls from height L/2 to height 0 when released. The velocity of COM when it reaches bottom is:

A √(gL/2)

B √(gL)

C √(2gL)

D √(gL/4)

Correct Answer: A. √(gL/2)

EXPLANATION

For rotating rod, energy: mg(L/2) = (1/2)Iω² + (1/2)mv². With I = mL²/3 about pivot, v_cm = √(gL/2)

Test

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