MHT CET · Physics · Center of Mass Momentum and Collision
A uniform rod \(\mathrm{AB}\) of mass ' \(\mathrm{m}^{\prime}\) and length ' \(\ell^{\prime}\) is at rest on a smooth horizontal surface. An impulse ' \(\mathrm{P}^{\prime}\) is applied to the end \(\mathrm{B}\). The time taken by the rod to turn through a right angle is
- A \(\frac{\pi}{12} \frac{\mathrm{m} \ell}{\mathrm{P}}\)
- B \(\frac{\pi \mathrm{P}}{\mathrm{m} \ell}\)
- C \(2 \pi \frac{\mathrm{m} \ell}{\mathrm{P}}\)
- D \(2 \frac{\pi \mathrm{P}}{\mathrm{m} \ell}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{12} \frac{\mathrm{m} \ell}{\mathrm{P}}\)
Step-by-step Solution
Detailed explanation
\(L = I \omega\) \(P \frac{\ell}{2} = \frac{1}{12} m \ell^2 \omega\)
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