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(A) The time taken for a body to roll down an inclined plane of length $L$ and angle of inclination $	heta$ is given by the formula: $t = \sqrt{\frac

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$1:1$

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(A) The time taken for a body to roll down an inclined plane of length $L$ and angle of inclination $	heta$ is given by the formula: $t = \sqrt{\frac{2L(1 + k^2/R^2)}{g \sin 	heta}}$,where $k$ is the radius of gyration and $R$ is the radius of the body.For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $k^2 = \frac{1}{2}R^2$,which gives $k^2/R^2 = 1/2$.For a disc,the moment of inertia $I = \frac{1}{2}MR^2$,so $k^2 = \frac{1}{2}R^2$,which gives $k^2/R^2 = 1/2$.Since both the solid cylinder and the disc have the same moment of inertia factor $(k^2/R^2 = 1/2)$,the time taken for both to reach the bottom is the same.Therefore,the ratio of their times is $1:1$.

$A$ solid cylinder and a disc of same radii are allowed to roll down a rough inclined plane from the top of a plane. The ratio of their times taken to reach the bottom of the inclined plane is

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