A string is wrapped around a wheel of radius ‘r’. The axis of the wheel is horizontal and its M.I. about the axis is I. Weight ‘mg’ is tied to free end of the string which is released to fall down from rest position. The angular velocity of the wheel after it has fallen through distance ‘h’ will be

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${(\frac{2 g h}{I + m r^{2}})}^{\frac{1}{2}}$

${(\frac{2 m g h}{I + m r^{2}})}^{\frac{1}{2}}$

$\frac{2 m g h}{I + 2 m r^{2}}$

$\sqrt{2 g h}$

answer is B.

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Detailed Solution

The change in potential energy of the weight will be converted into the kinetic energy of the wheel plus weight

$m g h = \frac{1}{2} m ϑ^{2} + \frac{1}{2} I ω^{2}$

= $\frac{1}{2} m r^{2} ω^{2} + \frac{1}{2} I ω^{2}$

$m g h = \frac{1}{2} ω^{2} (m r^{2} + I)$

$ω = \sqrt{\frac{2 m g h}{I + m r^{2}}}$

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