Question

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

• A. At the top of the circle
• B. At the bottom of the circle
• C. Halfway Down
• D. None of the above

Answer 1

The correct option is B At the bottom of the circle

$\begin{array}{c}T=\frac{m{v}^2}{r}+mg\cos x\\ 52=\frac{2\times 16}{1}+2\times 10\cos x\\ \cos x=1\\ x=0\\ So,particlewillbeatlowermostposition\end{array}.$

Hence,

option $\left(B\right)$ is correct answer.