As shown in the figure, a bob of mass m is tied by a mass less string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the speed of the bob will be

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$\sqrt{\frac{3}{2 gh}}$

$\sqrt{\frac{3}{4 gh}}$

$\sqrt{\frac{4 gh}{3}}$

$\sqrt{\frac{2 gh}{3}}$

answer is C.

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Detailed Solution

From the law of conservation of mechanical energy,

$mgh = \frac{1}{2} {mv}^{2} + \frac{1}{2} {Iω}^{2}$

As there is no slipping, so v = $ω$ r
$\begin{array}{l} mgh = \frac{1}{2} m (ωr)^{2} + \frac{1}{2} \frac{{mr}^{2}}{2} ω^{2} [∵ For disk, I = \frac{1}{2} {mr}^{2}] \\ mgh = \frac{3}{4} {mω}^{2} r^{2} \Rightarrow ω = \frac{1}{r} \sqrt{\frac{4}{3} gh} \end{array}$