A simple pendulum bob in a position when the string is vertical, is projected with such a velocity that its velocity in the horizontal position becomes the half. In the horizontal position ,net force on the bob is :

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$\frac{5}{3} m g$

$3 m g$

$\frac{\sqrt{13}}{3} m g$

$\frac{3}{5} m g$

answer is D.

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Detailed Solution

$\frac{1}{2} m {v_{B}}^{2} = \frac{1}{2} m {(\frac{v_{B}}{2})}^{2} + m g r, {v_{B}}^{2} = \frac{8}{3} g r, v_{A} = \frac{1}{2} v_{B} = \frac{1}{2} \sqrt{\frac{8}{3} g r} = \sqrt{\frac{2}{3} g r}$

$T_{A} = m g \cos 90^{0} + \frac{m {v^{2}}_{A}}{r} = \frac{m}{r} . \frac{2}{3} g r = \frac{2}{3} m g$

$n e t f o r c e = \sqrt{{(m g)}^{2} + {(\frac{2}{3} m g)}^{2}} = m g \frac{\sqrt{13}}{3}$

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