A stone is tied to a string of length ' $l$ ' and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed ' $u$ '. The magnitude of change in velocity as it reaches a position where the string is horizontal ( $g$ being acceleration due to gravity)is:

• Correct answer
  A

  $\sqrt{2\left(u^2-g l\right)}$
• B

  $\sqrt{u^2-g l}$
• C

  $u-\sqrt{u^2-2 g l}$
• D

  $\sqrt{2 g l}$

Solution

According to the question Total energy at $\mathrm{A}=$ Total energy at $\mathrm{B}$ $\begin{array}{l} \Rightarrow \frac{1}{2} m u^2 =\frac{1}{2} m v^2+m g l \\ \Rightarrow v^2 =\sqrt{u^2-2 g l} \end{array}$ $\therefore$ Change in magnitude of velocity $=\sqrt{u^2+v^2}=\sqrt{2\left(u^2-g l\right)}$

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