When a solid cylinder rolls down a rough inclined plane without slipping its acceleration is $' a_{1}'$ . When the same cylinder moves down a perfectly frictionless inclined plane having same inclination with horizontal, its acceleration is $' a_{2}'$ . Then $\frac{a_{1}}{a_{2}}$ is

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$\frac{3}{5}$

$\frac{2}{5}$

$\frac{2}{3}$

$\frac{4}{3}$

answer is C.

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Detailed Solution

For rolling without slipping, $a_{1} = \frac{gsinθ}{(1 + \frac{\frac{1}{2} {mR}^{2}}{{mR}^{2}})} = \frac{2}{3} gsinθ$

For sliding without rolling, $a_{2} = gsinθ$

$∴ \frac{a_{1}}{a_{2}} = \frac{2}{3}$

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