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(D) The velocity $V$ of a body rolling down an inclined plane of height $H$ without slipping is given by the formula: $V = \sqrt{\frac{2gH}{1 + k^2/R^

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$\sqrt{\frac{14}{15}}$

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(D) The velocity $V$ of a body rolling down an inclined plane of height $H$ without slipping is given by the formula: $V = \sqrt{\frac{2gH}{1 + k^2/R^2}}$,where $k$ is the radius of gyration.For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $k^2/R^2 = 1/2$.For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$,so $k^2/R^2 = 2/5$.The ratio of velocities is given by: $\frac{V_{	ext{cylinder}}}{V_{	ext{sphere}}} = \sqrt{\frac{1 + k_{	ext{sphere}}^2/R^2}{1 + k_{	ext{cylinder}}^2/R^2}}$.Substituting the values: $\frac{V_{	ext{cylinder}}}{V_{	ext{sphere}}} = \sqrt{\frac{1 + 2/5}{1 + 1/2}} = \sqrt{\frac{7/5}{3/2}} = \sqrt{\frac{7}{5} 	imes \frac{2}{3}} = \sqrt{\frac{14}{15}}$.

$A$ solid cylinder and a solid sphere,having same mass $M$ and radius $R$,roll down the same inclined plane from the top without slipping. They start from rest. The ratio of the velocity of the solid cylinder to that of the solid sphere,with which they reach the ground,will be:

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$A$ solid sphere at rest rolls down an inclined plane of vertical height $h$ without sliding. Its speed on reaching the bottom of the plane is ($g=$ acceleration due to gravity).

Suppose a body of mass $M$ and radius $R$ is allowed to roll on an inclined plane without slipping from its topmost point $A$. The acceleration of the body down the plane is given by (where $\beta = 1 + \frac{I}{MR^2}$):

$A$ solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
$(a)$ Will it reach the bottom with the same speed in each case?
$(b)$ Will it take longer to roll down one plane than the other?
$(c)$ If so,which one and why?

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 \,cm$. If the cylinder rolls without slipping, then the speed when it reaches the bottom is (in $\,m/s$)

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