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Mechanics

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

36 Q 9 Topics Take Test

Difficulty: All Easy Medium Hard 1–10 of 36

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.1 Easy Mechanics

An elevator accelerates upward at 2 m/s². A person of mass 60 kg stands in it. What is the normal force exerted by the floor on the person? (g = 10 m/s²)

A 600 N

B 660 N

C 720 N

D 540 N

Correct Answer: C. 720 N

EXPLANATION

N - mg = ma (upward acceleration), so N = m(g + a) = 60(10 + 2) = 60 × 12 = 720 N

Test

Q.2 Easy Mechanics

A body of mass 2 kg moves in a circular path of radius 0.5 m with constant speed. If the centripetal acceleration is 8 m/s², what is the centripetal force required?

A 16 N

B 8 N

C 4 N

D 2 N

Correct Answer: A. 16 N

EXPLANATION

Centripetal force F = ma_c = 2 × 8 = 16 N. This is the net force directed toward the center of circular motion.

Test

Q.3 Easy Mechanics

A projectile is launched at an angle of 45° with initial velocity 20 m/s. Neglecting air resistance, what is the maximum height reached? (g = 10 m/s²)

A 10 m

B 20 m

C 5 m

D 15 m

Correct Answer: A. 10 m

EXPLANATION

At 45°, vertical component = 20sin(45°) = 10√2 m/s. Using v² = u² - 2gh, at max height v=0: h = u²/(2g) = (10√2)²/(2×10) = 200/20 = 10 m

Test

Q.4 Easy Mechanics

A particle undergoes SHM with amplitude A = 0.5 m and angular frequency ω = 4 rad/s. What is its maximum velocity?

A 1 m/s

B 2 m/s

C 4 m/s

D 8 m/s

Correct Answer: B. 2 m/s

EXPLANATION

Maximum velocity in SHM = ωA = 4 × 0.5 = 2 m/s

Test

Q.5 Easy Mechanics

A particle experiences three forces: F₁ = 3i N, F₂ = 4j N, and F₃ = -3i - 4j N. What is the net force magnitude?

A 0 N

B 5 N

C 7 N

D 10 N

Correct Answer: A. 0 N

EXPLANATION

Net force = (3-3)i + (4-4)j = 0i + 0j = 0 N. The magnitude is 0 N.

Test

Q.6 Easy Mechanics

A block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²)

A 5 m/s

B 10 m/s

C 15 m/s

D 20 m/s

Correct Answer: B. 10 m/s

EXPLANATION

Using energy conservation: mgh = ½mv². v = √(2gh) = √(2×10×5) = √100 = 10 m/s

Test

Q.7 Easy Mechanics

What is the moment of inertia of a uniform disc of mass M and radius R about an axis perpendicular to the disc passing through its center?

A MR²

B ½MR²

C ⅓MR²

D ¼MR²

Correct Answer: B. ½MR²

EXPLANATION

The moment of inertia of a uniform disc about its central perpendicular axis is I = ½MR²

Test

Q.8 Easy Mechanics

A particle moves with constant velocity v = 5 m/s in a circular path of radius 10 m. What is its centripetal acceleration?

A 0.5 m/s²

B 2 m/s²

C 2.5 m/s²

D 5 m/s²

Correct Answer: C. 2.5 m/s²

EXPLANATION

Centripetal acceleration = v²/r = 25/10 = 2.5 m/s²

Test

Q.9 Easy Mechanics

A block of mass 5 kg is placed on a frictionless inclined plane at 30°. What is the acceleration down the plane? (g = 10 m/s²)

A 5 m/s²

B 8.66 m/s²

C 5√3 m/s²

D 2.5 m/s²

Correct Answer: A. 5 m/s²

EXPLANATION

a = g sin(30°) = 10 × 0.5 = 5 m/s²

Test

Q.10 Easy Mechanics

A body is projected vertically upward with velocity 40 m/s. If g = 10 m/s², what is the maximum height reached?

A 40 m

B 80 m

C 160 m

D 20 m

Correct Answer: B. 80 m

EXPLANATION

Using v² = u² - 2gh, at maximum height v = 0. So 0 = 1600 - 2(10)h, h = 80 m

Test

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