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Question

The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height $$h$$ from rest without sliding is


A
(g5)gh
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B
gh
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C
(43)gh
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D
(107)gh
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Solution

The correct option is A $$\sqrt{\left (\dfrac{g}{5} \right) gh}$$
For solid sphere   $$I= \dfrac{2}{5}mR^2$$
Work- energy theorem,    $$mgh= \dfrac{1}{2}Iw^2+ \dfrac{1}{2}mv^2$$
 $$mgh= \dfrac{1}{2}\times\dfrac{2}{5}mR^2w^2+ \dfrac{1}{2}mv^2$$                     where   $$v=Rw$$      (due to pure rolling)
$$\implies v=\sqrt{\dfrac{10}{7}gh}$$

Physics

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