Question

A stone tide 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 change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt{x\left(u^2-gL\right)}$. The value of $x$ is

• A. 3
• B. 2
• C. 1
• D. 5

Answer 1

Answer: (B)

$v=\sqrt{u^2-2gL}$
$\Delta v=\sqrt{u^2+v^2}$
$\Delta v=\sqrt{u^2+v^2-2gL}$
$\Delta v=\sqrt{2u^2-2gL}$
$\Delta v=\sqrt{2\left(u^2-gL\right)}x=2$