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

4 m / s^{1}

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2 m / s^{1}

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2.5 m / s^{1}

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None of these

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Solution

The correct option is C $4 m / s^{1}$
From $F B D$
$T_{1} cos 30^{0} + T_{2} cos 45^{0} = m g$
$T_{1} sin 30^{0} + T_{2} sin 45^{0} = \frac{m v^{2}}{r}$
$T_{1} sin 30^{0} - T_{1} cos 30^{0} = \frac{m v^{2}}{R} - m g$ $∵ {sin 45^{0} = cos 45^{0}}$
$T_{1} = \frac{\frac{m v^{2}}{R} - m g}{[\frac{1}{2} - \frac{\sqrt{3}}{2}]}$

as it a string tension, So it can't be negative
$T_{1} > 0$ i.e $\frac{m v^{2}}{R} - m g \geq 0$
$\frac{v_{2}}{r} - g \geq 0$
$v \geq \sqrt{g R}$
$V_{m i n} \sqrt{g r} = \sqrt{10 \times 1.6} = \sqrt{16} = 4 m s^{- 1}$
Hence (A) is the correct options.

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