A small object of uniform density rolls up a curved surface with in an initial velocity ' $v$ '. It

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A small object of uniform density rolls up a curved surface with in an initial velocity ' $v$ '. It reaches upto a maximum height of $\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}$ with respect to the vertical position. The object is\

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(B) The kinetic energy of the rolling object is converted into potential energy at height $h\left(=\frac{3 v^{2}}{4 g}ight)$\ So by the law of conservation of mechanical energy, we have\ $\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{Mgh}$\ $\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{v}}{\mathrm{R}}ight)^{2}=\mathrm{Mg}\left(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}ight)\left(\because \omega=\frac{\mathrm{v}}{\mathrm{R}}ight)$\ $\frac{1}{2} \mathrm{I} \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{3}{4} \mathrm{Mv}^{2}-\frac{1}{2} \mathrm{Mv}^{2}$\ $\frac{1}{2} \mathrm{I} \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{1}{4} \mathrm{Mv}^{2}$ or $\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}$\ Hence, the object is disc.\