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Question

A small object of uniform density rolls up a curved surface with an initial velocity $$v$$. It reaches up to a maximum height of $$\dfrac{3v^2}{4g}$$ w.r.t. the initial position. The object is a:
120443_3ccf5dda7fd243e79cc93e673c8e0c91.png


A
ring
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B
solid sphere
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C
hollow sphere
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D
disc
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Solution

The correct option is C disc
By energy balance we get
$$\displaystyle\frac{1}{2}mv^2 + \displaystyle\frac{1}{2}I\omega^2=mgh$$
or
$$\displaystyle\frac{1}{2}mv^2+\displaystyle\frac{1}{2}I(\displaystyle\frac{v}{R})^2=mg\displaystyle\frac{3v^2}{4g}$$
or
$$I=\displaystyle\frac{mr^2}{2}$$
This is the moment of inertia for a solid disc. Thus option D is correct.

Physics

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