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

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

Answer 1

Answer: (A)

Here $v^2-u^2=-2gl$ ....(i)

$v^2=u^2-2gl$

Since the velocities are mutually perpendicular, change in velocity

$△v=\sqrt{u^2+v^2}$ ....(ii)

$=\sqrt{u^2+u^2-2gl}$

(substituting the value of $v^2$ from (i))

or $△v=\sqrt{2(u^2-gl)}$