A rod of length $l$ is held vertically stationary with its lower end located at a point $P$, on the horizontal plane. When the rod is released to topple about $P$, the velocity of the upper end of the rod with which it hits the ground is

• A

  $\sqrt{\frac{g}{l}}$
• Correct answer
  B

  $\sqrt{3 g l}$
• C

  $3 \sqrt{\frac{g}{l}}$
• D

  $\sqrt{\frac{3 g}{l}}$

Solution

In this process potential energy of the metre stick will be converted into rotational kinetic energy. PE of metre stick $=\frac{m g l}{2}$ Because its centre of gravity lies at the middle of the rod.
Rotational kinetic energy $E=\frac{1}{2} I \omega^2$ $I=$ moment of inertia of metre stick about point $A=\frac{m l^2}{3} \text {. }$ By the law of conservation of energy $\begin{aligned} m g\left(\frac{l}{2}\right) & =\frac{1}{2} I \omega^2 \\ & =\frac{1}{2} \frac{m l^2}{3}\left(\frac{v_B}{l}\right)^2 \end{aligned}$ By solving, we get $v_B=\sqrt{3 g l}$

Hence, the correct answer is option: 'B'.