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A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle of inclination 30°. The centre of mass of the cylinder has speed of 4 m/s. The distance travelled by the cylinder on the inclined surface will be, $[take g = 10 m / s^{2}]$

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2.4 m

2.2 m

1.6 m

1.2 m

answer is A.

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Detailed Solution

$Since v = \sqrt{\frac{2 gh}{1 + \frac{K^{2}}{R^{2}}}}$

$v^{2} = \frac{2 gh}{1 + \frac{1}{2}} [\frac{K^{2}}{R^{2}} = \frac{1}{2} for solid cylinder] 2 gh = 4^{2} \times \frac{3}{2} h = \frac{12}{10} = 1.2 m Now x = \frac{h}{\sin 30 °} = \frac{1.2}{\frac{1}{2}} = 2.4 m x = 2.4 m$

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