Question

A thin rod of mass $m$ and length $2l$ is made to rotate about an axis passing through its centre and perpendicular to it. If its angular velocity changes from $0$ to $\omega$ in time $t$, the torque acting on it is

• A. $\frac{ml^2 \omega}{12 t}$
• B. $\frac{ml^2 \omega}{3 t}$
• C. $\frac{ml^2 \omega}{t}$
• D. $\frac{4ml^2 \omega}{3 t}$

Answer 1

Answer: (B)

$\frac{ml^2 \omega}{3 t}$

Answer 2

since $\tau=l \alpha$ so,$\tau=\left[\frac{m(2l)^2}{12}\right]\left(\frac{\omega}{t}\right) \,or\,\tau =\frac{m \times 4 l^2 \times \omega }{12 \times t}$ or $\tau=\frac{4ml^2 \omega}{12t}=\left(\frac{ml^2\omega}{3t}\right)$