Question

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 the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is

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

Answer 1

The Correct Option is A

Key Idea : When stone reaches a position where string is horizontal, it attains the energy partially as kinetic and partially as potential. When stone is at its lowest position, it has only kinetic energy, given by $K=\frac{1}{2} m u^{2}$

At the horizontal position, it has energy $ E = U + K = \frac{1}{2} mu'^2 + mgl$

According to conservation of energy, $K =E $ $\therefore \frac{1}{2} m u^{2} =\frac{1}{2} m u^{\prime 2}+m g l$ or $\frac{1}{2} m u^{\prime 2}=\frac{1}{2} m u^{2}-m g l$ or $u^{\prime 2} =u^{2}-2 g l $ or $u^{\prime} =\sqrt{u^{2}-2 g l} \ldots$(i) So, the magnitude of change in velocity

$|\Delta \vec{ u }| =|\overrightarrow{ u }|=\sqrt{u^{\prime 2}+u^{2}+2 u^{\prime} u \cos 90^{\circ}} $ $|\Delta \overrightarrow{ u }| =\sqrt{u^{\prime 2}+u^{2}} $ $=\sqrt{2\left(u^{2}-g L\right)} $ [from E (i)]