From an inclined plane,a sphere,a disc,a ring | vedclass

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(C) The time taken by a rolling body to reach the bottom of an inclined plane is given by:$t = \frac{1}{\sin 	heta} \sqrt{\frac{2 h}{g} \left(1 + \fr

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Sphere,disc,shell,ring

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(C) The time taken by a rolling body to reach the bottom of an inclined plane is given by:$t = \frac{1}{\sin 	heta} \sqrt{\frac{2 h}{g} \left(1 + \frac{K^2}{R^2}ight)}$Since $	heta$,$h$,and $g$ are constant for all bodies,the time taken is proportional to $\sqrt{1 + \frac{K^2}{R^2}}$.We calculate the values of $\frac{K^2}{R^2}$ for each object:For a ring: $\frac{K^2}{R^2} = 1$For a disc: $\frac{K^2}{R^2} = \frac{1}{2} = 0.5$For a solid sphere: $\frac{K^2}{R^2} = \frac{2}{5} = 0.4$For a spherical shell: $\frac{K^2}{R^2} = \frac{2}{3} \approx 0.67$Comparing these values,we get: $\left(\frac{K^2}{R^2}ight)_{	ext{sphere}} Therefore,the order of time taken is: $t_{	ext{sphere}} < t_{	ext{disc}} < t_{	ext{shell}} < t_{	ext{ring}}$.

From an inclined plane,a sphere,a disc,a ring,and a spherical shell of the same radius are rolled without slipping from the same height simultaneously. The order of their reaching at the base will be:

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