JEE (Main + Adv) · Physics
Rotational Motion
15 practice questions with full step-by-step solutions — free, no sign-up.
This chapter has
Rotational Motion — solved practice questions
8 JEE Physics questions with step-by-step solutions. Attempt each, then reveal the answer.
- Q1easy
The moment of inertia of a uniform solid sphere of mass M and radius R about a diameter is:
- A(1/2)MR^2
- B(2/5)MR^2
- CMR^2
- D(2/3)MR^2
Show answer & solution
Correct answer: (B) (2/5)MR^2
The standard result for a solid sphere about a diameter is I = (2/5)MR^2.
- Q2medium
A solid sphere rolls without slipping down an incline of angle 30 degrees (g = 10 m/s^2). Its linear acceleration is:
- A5 m/s^2
- B2.5 m/s^2
- C25/7 m/s^2
- D10 m/s^2
Show answer & solution
Correct answer: (C) 25/7 m/s^2
For rolling without slipping, a = g sin(theta)/(1 + I/MR^2) = 10*0.5/(1 + 2/5) = 5/1.4 = 25/7 = 3.57 m/s^2.
- Q3medium
A solid sphere rolls without slipping. The fraction of its total kinetic energy that is rotational is:
- A2/7
- B5/7
- C1/2
- D1/3
Show answer & solution
Correct answer: (A) 2/7
Rotational KE = (1/2)(2/5 MR^2)(v/R)^2 = (1/5)Mv^2 and total = (7/10)Mv^2, so the fraction is (1/5)/(7/10) = 2/7.
- Q4easy
A ballet dancer spins with her arms outstretched. When she pulls her arms in, reducing her moment of inertia to half, her angular velocity:
- Ahalves
- Bstays the same
- Cbecomes one-fourth
- Ddoubles
Show answer & solution
Correct answer: (D) doubles
With no external torque, angular momentum L = I*omega is conserved. Halving I doubles omega.
- Q5easy
A uniform disc of mass 2 kg and radius 0.5 m rotates about its central axis. Its moment of inertia is:
- A0.5 kg m^2
- B0.25 kg m^2
- C1 kg m^2
- D0.125 kg m^2
Show answer & solution
Correct answer: (B) 0.25 kg m^2
For a disc about its central axis, I = (1/2)MR^2 = (1/2)(2)(0.5)^2 = 0.25 kg m^2.
- Q6medium
A flywheel starts from rest and gains angular speed under a constant angular acceleration of 2 rad/s^2. The angle turned through in the first 5 s is:
- A10 rad
- B50 rad
- C25 rad
- D5 rad
Show answer & solution
Correct answer: (C) 25 rad
theta = omega0*t + (1/2)*alpha*t^2 = 0 + (1/2)(2)(25) = 25 rad.
- Q7easy
Two point masses of 2 kg and 3 kg are placed at x = 0 and x = 5 m on a light rod. The centre of mass lies at:
- A2.5 m
- B2 m
- C3.5 m
- D3 m
Show answer & solution
Correct answer: (D) 3 m
x_cm = (m1x1 + m2x2)/(m1 + m2) = (2*0 + 3*5)/(2+3) = 15/5 = 3 m.
- Q8medium
A ring, a disc, and a solid sphere of the same mass and radius are released from rest at the top of the same incline and roll down without slipping. Which reaches the bottom first?
- ASolid sphere
- BDisc
- CRing
- DAll together
Show answer & solution
Correct answer: (A) Solid sphere
Acceleration a = g sin(theta)/(1 + I/MR^2) is largest for the smallest I/MR^2. The solid sphere has 2/5, the smallest, so it arrives first.
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