Home › Subjects › JEE Physics

JEE Physics

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

434 Q 9 Topics Take Test

Difficulty: All Easy Medium Hard 401–410 of 434

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.401 Medium Mechanics

A particle moving in circular path has centripetal acceleration a_c and tangential acceleration a_t. The magnitude of total acceleration is:

A a_c + a_t

B √(a_c² - a_t²)

C √(a_c² + a_t²)

D |a_c - a_t|

Correct Answer: C. √(a_c² + a_t²)

EXPLANATION

Centripetal and tangential accelerations are perpendicular. Total acceleration = √(a_c² + a_t²)

Test

Q.402 Medium Mechanics

A particle undergoes simple harmonic motion with amplitude A and period T. The maximum velocity is:

A AT

B 2πA/T

C A/T

D πA/T

Correct Answer: B. 2πA/T

EXPLANATION

In SHM, v_max = ωA = (2π/T)A = 2πA/T

Test

Q.403 Medium Mechanics

A wheel rotating at 100 rpm is brought to rest by applying brakes for 10 seconds. The angular deceleration is:

A π/3 rad/s²

B π/6 rad/s²

C 2π/3 rad/s²

D π rad/s²

Correct Answer: A. π/3 rad/s²

EXPLANATION

Initial ω = 100×2π/60 = 10π/3 rad/s. Final ω = 0. α = Δω/Δt = (10π/3)/10 = π/3 rad/s²

Test

Q.404 Medium Mechanics

A block slides down an inclined plane of angle θ and length L. If friction coefficient is μ, the velocity at bottom is:

A √(2gL(sin θ + μ cos θ))

B √(2gL(sin θ - μ cos θ))

C √(gL sin θ)

D √(2gL sin θ)

Correct Answer: B. √(2gL(sin θ - μ cos θ))

EXPLANATION

Net acceleration = g(sin θ - μ cos θ). Using v² = 2aL, v = √(2gL(sin θ - μ cos θ))

Test

Q.405 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.406 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.407 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.408 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.409 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.410 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

1…39 40 41 42 43 44

All Subjects & Topics

Govt. Exams

Entrance Exams

Teacher Exams

Subjects

Quick Links

JEE Physics