A rolling wheel of $12 kg$ is on an inclined plane at position P and connected to a mass of $3 kg$ through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface.
The velocity of center of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $\frac{1}{2} \sqrt{x g h} m / s$ . The value of $x$ is __________.

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answer is 2.67.

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Detailed Solution

Net loss in PE = Gain in KE
$12 gh - 3 gh = \frac{1}{2} 3 v^{2} + \frac{1}{2} 12 v^{2} + \frac{1}{2} [12 r^{2}] {(\frac{v}{r})}^{2} 9 gh = \frac{1}{2} [3 + 12 + 12] v^{2} v^{2} = \frac{2 gh}{3} \Rightarrow v = \frac{1}{2} \sqrt{\frac{8}{3} gh} x = \frac{8}{3} = 2.67$

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