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

• A. 3
• B. 2
• C. 1
• D. 5

Answer 1

Answer: (B)

2

Answer 2

$\overrightarrow v=\sqrt{u^2−2gL\hat j}$

$\overrightarrow u→=u\hat i$

$∴|\overrightarrow v−\overrightarrow u|=\sqrt{(u^2−2gl)+u^2}$

=$\sqrt{2u^2−2gl}$

$∴ x = 2$