A wheel having angular momentum $2 \pi \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}$ about its vertical axis,…

A wheel having angular momentum $2 \pi \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}$ about its vertical axis, rotates at the rate of $60 \mathrm{rpm}$ about this axis, The torque which can stop the wheel's rotation in 30 sec would be
  1. $\frac{2 \pi}{15} \mathrm{Nm}$
  2. $\frac{\pi}{18} \mathrm{Nm}$
  3. $\frac{\pi}{12} \mathrm{Nm}$
  4. $\frac{\pi}{15} \mathrm{Nm}$

Solution

or $\begin{aligned} \tau \times \Delta \mathrm{t} &=\mathrm{L}_{0} &\left\{\because \text { since } \mathrm{L}_{f}=0\right\} \\ \tau \times 30 &=2 \pi & & \\ \tau &=\frac{\pi}{15} \mathrm{~N}-\mathrm{m} & \end{aligned}$

Asked in: JEE Mains - Rotational Motion - Test 2

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