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A solid cylinder starts pure rolling from top of a fixed inclined plane of slant length 60cm and inclination $30^{0} .$ Velocity of centre of mass when the cylinder reaches the bottom is equal to $({g = 10m/s}^{2})$

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1 m/s

3 m/s

2 m/s

4 m/s

answer is C.

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Detailed Solution

$v = \sqrt{\frac{2 g l \sin θ}{1 + \frac{K^{2}}{R^{2}}}} = \sqrt{\frac{2×10×0.6× \frac{1}{2}}{1+ \frac{1}{2}}} =2m/s$

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