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A ring is rolling without sliping on an inclined plane . Find its velocity (in m/sec) after moving on inclined plane by distance 10 m , releasing from rest . Assume \theta = 30 \degree of inclined of g = 9.8 m/s

Option: 1

5


Option: 2

7


Option: 3

4


Option: 4

3


Answers (1)

Condition for pure rolling on inclined plane -

\mu _{s}\geq \frac{\tan \Theta }{1+\frac{R^{2}}{K^{2}}}

- wherein

\mu _{s} = limiting cofficient of friction

 

 

By energy conservation 

mgh = \frac{1}{2}mv^2 + \frac{1}{2}IW^2\\\\mgh = \frac{1}{2}mv^2 + \frac{1}{2}mK^2 \left ( \frac{V}{R} \right )^2\\\\ mgh = \frac{1}{2}mv^2 \left [ 1+\frac{K^2}{R^2} \right ]\\\\v=\sqrt{\frac{2gh}{1+\frac{k^2}{R^2}}}\\\\ v = \sqrt{\frac{2\times 9.8\times 10\times \sin 30 \degree}{1+1}} \Rightarrow v = 7 m/s

 

Posted by

Sumit Saini

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