A $2 k g$ at the end of a string $1 m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4 m / s e c$ . The tension is the string will be $52 N$ , when the stone is -

At the top of the circle

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At the bottom of the circle

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Halfway Down

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None of the above

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Solution

The correct option is B At the bottom of the circle
$\begin{matrix} T = \frac{m v^{2}}{r} + m g cos x 52 = \frac{2 \times 16}{1} + 2 \times 10 cos x cos x = 1 x = 0 S o, p a r t i c l e w i l l b e a t l o w e r m o s t p o s i t i o n \end{matrix} .$
Hence,
option $(B)$ is correct answer.

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