A solid sphere of uniform density and radius | vedclass

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(D) The gravitational potential $V$ at any point $P$ due to the solid sphere with cavities is given by the principle of superposition: $V = V_{	ext{l

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) The gravitational potential $V$ at any point $P$ due to the solid sphere with cavities is given by the principle of superposition: $V = V_{	ext{large sphere}} - V_{	ext{cavity A}} - V_{	ext{cavity B}}$.$1$. At the origin $O(0, 0, 0)$,the gravitational field $\vec{E}$ due to the large sphere is zero. The fields due to the two cavities are equal in magnitude and opposite in direction $(\vec{E}_A = -\vec{E}_B)$,so the net field at the origin is zero. Thus,option $(A)$ is correct.$2$. The gravitational potential at any point $(x, y, z)$ due to a sphere with a cavity is symmetric about the axis passing through the centres of the cavities. For the circle ${y^2} + {z^2} = R^2$ in the plane $x = 0$,the distance from any point on the circle to the centres $A(-2, 0, 0)$ and $B(2, 0, 0)$ is the same. Specifically,for any point $(0, y, z)$ on the circle,the distance to $A$ is $\sqrt{(-2-0)^2 + y^2 + z^2} = \sqrt{4 + R^2}$ and the distance to $B$ is $\sqrt{(2-0)^2 + y^2 + z^2} = \sqrt{4 + R^2}$. Since the distances are equal,the potentials due to the cavities are equal,and the potential due to the large sphere is constant on this circle. Thus,the total potential is constant on the circles ${y^2} + {z^2} = 4$ and ${y^2} + {z^2} = 36$. Options $(B)$ and $(C)$ are also correct.Therefore,the correct option is $(D)$.

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