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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

A solid disc of radius \(20 \,{cm}\) and mass \(10\, {kg}\) is rotating with an angular velocity of \(600\, {rpm}\), about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in \(10\, {s}\, ......\, \pi \times \,10^{-1}\, {Nm}\).

  1. A \(1\)
  2. B \(2\)
  3. C \(3\)
  4. D \(4\)
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Answer & Solution

Correct Answer

(D) \(4\)

Step-by-step Solution

Detailed explanation

\(\tau=\frac{\Delta L}{\Delta t}=\frac{I\left(\omega_{f}-\omega_{i}\right)}{\Delta t}\) \(\tau=\frac{\frac{m R^{2}}{2} \times[0-\omega]}{\Delta t}\) \(=\frac{10 \times\left(20 \times 10^{-2}\right)^{2}}{2} \times \frac{600 \times \pi}{30 \times 10}\)…
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