A solid cylinder of mass M and radius R rolls without slipping down an inclined plane making an angle $θ$ with the horizontal. Then its acceleration is

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$\frac{1}{3} g \sin θ$

$\frac{2}{3} gsin θ$

$\frac{2}{5} g \sin θ$

$\frac{2}{7} g \sin θ$

answer is B.

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Detailed Solution

Acceleration of a rolling body down an inclined plane is

a = $\frac{g \sin θ}{1 + \frac{k^{2}}{R^{2}}}$

For solid cylinder, $k^{2} = \frac{R^{2}}{2}$

$a = \frac{g \sin θ}{1 + \frac{1}{2}} = \frac{2}{3} g \sin θ$

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