Two wires $A C$ and $B C$ are tied at $C$ of small sphere of mass $5 k g$ , which revolves at a constant speed $v$ in the horizontal circle of radius $1.6 m$ . The minimum value of $v$ is?

3.01 m / s

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4.01 m / s

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8.2 m / s

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3.96 m / s

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Solution

The correct option is D $3.96 m / s$
From force diagram shown in figure,
$T_{1} cos 30^{o} + T_{2} cos 45^{o} = m g$ .........(i)
$T_{1} sin 30^{o} + T_{2} sin 45^{o} = \frac{m v^{2}}{r}$ .......(ii)
After solving eqs. (i) and (ii), we get
$T_{1} = \frac{m g - \frac{m v^{2}}{r}}{\frac{(\sqrt{3} - 1)}{2}}$
But $T_{1} > 0$
$∴ \frac{m g - \frac{m v^{2}}{r}}{\frac{\sqrt{3} - 1}{2}} > 0$
or $m g > \frac{m v^{2}}{r}$ or $v < \sqrt{r g}$
$∴ v_{m i n} = \sqrt{r g} = \sqrt{1.6 \times 9.8} = 3.96 m / s$ .

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