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JEE Main 2026 (6 Apr, Shift 1) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2026 (6 Apr, Shift 1)

Q6JEE Main 2026 · 6 APR · SHIFT 1Medium

Rotational Kinematics & Dynamics

A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are ______ and ______ respectively.

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Answer: (d)

  1. I=25MR2=25(5)(0.04)2=3.2×10−3 kg m2I = \frac{2}{5}MR^2 = \frac{2}{5}(5)(0.04)^2 = 3.2\times10^{-3}\text{ kg m}^2
  2. ω0=1200 rpm=40π rad/s\omega_0 = 1200\text{ rpm} = 40\pi\text{ rad/s}, α=40π10=4π rad/s2\alpha = \frac{40\pi}{10} = 4\pi\text{ rad/s}^2
  3. τ=Iα=3.2×10−3×4π=0.0128π N m\tau = I\alpha = 3.2\times10^{-3}\times4\pi = 0.0128\pi\text{ N m}
  4. θ=ω0t2=40π×102=200π rad=100\theta = \frac{\omega_0 t}{2} = \frac{40\pi\times10}{2} = 200\pi\text{ rad} = 100 rotations
  5. So the answer is (d) 0.0128π0.0128\pi N m, 100.
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