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  • Tell me? The speed of a homogeneous, solid sphere after rolling down in the inclined plane of vertical height h, from rest without sliding is

The speed of a homogeneous, solid sphere after rolling down in the inclined plane of vertical height h, from rest without sliding is

  • Option 1)

    \sqrt{\frac{10}{7}gh}

  • Option 2)

    \sqrt{gh}

  • Option 3)

    \sqrt{\frac{8}{3}gh}

  • Option 4)

    \sqrt{\frac{4}{3}gh}

 

Answers (1)

best_answer

As we learnt in 

Rolling of a body on an inclined plane -

a= \frac{g\sin \Theta }{1+\frac{K^{2}}{R^{2}}}

f= \frac{mg\sin \Theta }{1+\frac{R^{2}}{K^{2}}}

- wherein

K=Radius of gyration

\Theta = Angle of inclination

 

 V^{2} = \frac{2gh}{1+\frac{k^{2}}{r^{2}}}

K^{2} = \frac{2}{5}R^{2}

V^{2} = \frac{2gh}{1+\frac{2R^{2}}{5R^{2}}} = \frac{10}{7}gh

 

\therefore v=\sqrt{\frac{10}{7}}gh


Option 1)

\sqrt{\frac{10}{7}gh}

correct

Option 2)

\sqrt{gh}

Incorrect

Option 3)

\sqrt{\frac{8}{3}gh}

Incorrect

Option 4)

\sqrt{\frac{4}{3}gh}

Incorrect

Posted by

Aadil

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