JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform rod \(A B\) of mass \(2 \mathrm{~kg}\) and Length \(30 \mathrm{~cm}\) at rest on a smooth horizontal surface. An impulse of force \(0.2\ \mathrm{Ns}\) is applied to end \(B.\) The time taken by the rod to turn through at right angles will be \(\frac{\pi}{\mathrm{x}}\ \mathrm{s}\), where X = _______.
- A \(4\)
- B \(5\)
- C \(6\)
- D \(7\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
Impulse \(\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}\) \(\mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s}\) Angular impuls ( \(\overrightarrow{\mathrm{M}})\) \(\overrightarrow{\mathrm{M}}_{\mathrm{c}}=\int \tau \mathrm{dt}\)…
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