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NEET · Physics · STD 11 - 6. system of particles and rotational motion
A solid cylinder of mass \(2\; kg\) and radius \(4 \;\mathrm{cm}\) is rotating about its axis at the rate of \(3\; \mathrm{rpm}\). The torque required to stop after \(2 \pi\) revolutions is
- A \(2 \times 10^{-6} \; \mathrm{Nm}\)
- B \(2 \times 10^{-3} \; \mathrm{Nm}\)
- C \(12 \times 10^{-4}\; \mathrm{Nm}\)
- D \(2 \times 10^{6} \; \mathrm{Nm}\)
Answer & Solution
Correct Answer
(A) \(2 \times 10^{-6} \; \mathrm{Nm}\)
Step-by-step Solution
Detailed explanation
\(\theta= 2 \pi\) radian \(\omega_{0}=3 \mathrm{rpm} \Rightarrow \frac{2 \pi}{60}(3) \frac{\mathrm{rad}}{\mathrm{sec}}\)
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