Question

A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed $u$. The magnitude of the change in its velocity as it reaches a position, where the string is horizontal, is

• A. $ \sqrt{2({{u}^{2}}-gl)} $
• B. $ \sqrt{{{u}^{2}}-gl} $
• C. $ \sqrt{{{u}^{2}}-2gl} $
• D. $ \sqrt{2gl} $

Answer 1

Answer: (A)

From energy conservation $v^{2}=u^{2}-2 g L\,\,\,$....(i)

Now, since the two velocity vectors shown in figure are mutually perpendicular, hence the magnitude of change of velocity will be given by $|\Delta v |=\sqrt{u^{2}+v^{2}}$
Substituting value of $v^{2}$ from Eq.(i), we get
$|\Delta v |=\sqrt{u^{2}+u^{2}-2 g L}$
$=\sqrt{2\left(u^{2}-g L\right)}$