A solid cylinder of mass 2 kg and radius 50cm rolls up an inclined plane of angle inclination 30°. The centre of mass of cylinder has speed of 4 m/s. The distance travelled by the cylinder on the incline surface will be (Take g = 10 m/s²)

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1.2 m

2.4 m

2.2 m

1.6 m

answer is D.

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Detailed Solution

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

$v^{2} = \frac{2 gh}{1 + \frac{1}{2}}$
We know, $\frac{K^{2}}{R^{2}} = \frac{1}{2}$ for solid cylinder
$2 gh = 4^{2} x \frac{3}{2}$

$h = \frac{12}{10} = 1.2 m$

$Now, x = \frac{h}{\sin 30^{o}} = \frac{1.2}{1 / 2} = 2.4 m$

$x = 2.4 m$

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