Two discs having masses in the ratio 1:2 and | vedclass

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(D) For an object rolling down an inclined plane without slipping,the conservation of energy states that the potential energy at the top is equal to t

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$1:1$

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(D) For an object rolling down an inclined plane without slipping,the conservation of energy states that the potential energy at the top is equal to the total kinetic energy at the bottom.$mgh = K.E._{trans} + K.E._{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a disc,the moment of inertia $I = \frac{1}{2}mr^2$ and the rolling condition is $\omega = \frac{v}{r}$.Substituting these into the energy equation:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$v = \sqrt{\frac{4gh}{3}}$Since the final velocity $v$ depends only on the height $h$ and the acceleration due to gravity $g$,and is independent of the mass $m$ and radius $r$ of the disc,both discs will have the same linear velocity upon reaching the ground.Therefore,the ratio of their linear velocities is $1:1$.

Two discs having masses in the ratio $1:2$ and radii in the ratio $1:8$ roll down without slipping one by one from an inclined plane of height $h$. The ratio of their linear velocities on reaching the ground is ........

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