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(C) By the principle of conservation of energy,the loss in potential energy equals the gain in kinetic energy (translational + rotational).$mgh = \fra

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$120$

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(C) By the principle of conservation of energy,the loss in potential energy equals the gain in kinetic energy (translational + rotational).$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid cylinder,the moment of inertia $I = \frac{1}{2}mR^2$. Since the string unwinds without slipping,the condition $v = R\omega$ holds,which implies $\omega = \frac{v}{R}$.Substituting these into the energy equation:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$$mgh = \frac{3}{4}mv^2$$gh = \frac{3}{4}v^2$Given $g = 10\,ms^{-2}$ and $v = 4\,ms^{-1}$:$10h = \frac{3}{4}(4)^2$$10h = \frac{3}{4} 	imes 16$$10h = 12$$h = 1.2\,m = 120\,cm$.

$A$ solid cylinder is suspended symmetrically by two massless strings,as shown in the figure. The distance the cylinder should fall by unbinding the strings to achieve a speed of $4\,ms^{-1}$ is $........cm$. (Take $g=10\,ms^{-2}$)

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