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(D) The gravitational potential energy of a body is given by $U = mgh_{cm}$,where $h_{cm}$ is the height of the center of mass from the surface.In fig

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$\frac{h}{R} = 2\sqrt 2$

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(D) The gravitational potential energy of a body is given by $U = mgh_{cm}$,where $h_{cm}$ is the height of the center of mass from the surface.In figure $(A)$,the cone is standing on its base. The center of mass of a solid cone is at a height of $\frac{h}{4}$ from the base. Thus,$U_A = mg \left( \frac{h}{4} ight)$.In figure $(B)$,the cone is lying on its slant side. The center of mass is at a distance of $\frac{3h}{4}$ from the apex. Let $C$ be the center of mass. In the right-angled triangle formed by the apex $O$,the center of mass $C$,and the projection $M$ on the base,the height of the center of mass from the surface is $h_{cm} = \left( \frac{3h}{4} ight) \sin 	heta$.Since the potential energy does not change,$U_A = U_B$,so $\frac{h}{4} = \left( \frac{3h}{4} ight) \sin 	heta$.This gives $\sin 	heta = \frac{1}{3}$.We know that $	an 	heta = \frac{R}{h}$. Since $\sin 	heta = \frac{1}{3}$,the opposite side is $1$ and the hypotenuse is $3$. The adjacent side is $\sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$.Therefore,$	an 	heta = \frac{1}{2\sqrt{2}}$.Since $	an 	heta = \frac{R}{h}$,we have $\frac{R}{h} = \frac{1}{2\sqrt{2}}$,which implies $\frac{h}{R} = 2\sqrt{2}$.

$A$ solid cone is placed on a horizontal surface has height $h$,radius $R$ and apex angle $\theta$ as shown. If the gravitational potential energy of the cone does not change as the position of the cone is changed from figure $(A)$ to figure $(B)$,then,

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