Subtracting the contribution of the removed sphere from that of the entire sphere yields V = -(11GM)/(8R) + (3GM)/(8R) = -(8GM)/(8R) = -(GM)/(R). This corresponds to option (2).

JEE Main 2015 3. From a solid sphere of mass $M$ and radius $R$ , a spherical portion of radius $R/2$ is removed, as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$ , the potential at the centre of the cavity thus formed is ( $G = \text{gravitational constant}$ ) Figure description: A large shaded circular region represents a solid sphere of radius $R$ with its center marked by a small dot. Inside it, touching the right boundary of the large sphere, is an unshaded (white) circular region representing the removed cavity of radius $R/2$ . The center of this cavity is marked with a bold black dot, which is located at a distance of $R/2$ from the center of the original sphere. (1) $\frac{-GM}{2R}$ (2) $\frac{-GM}{R}$ (3) $\frac{-2GM}{3R}$ (4) $\frac{-2GM}{R}$

Answer: (2) $\frac{-GM}{R}$

Step-by-step solution

Step 1: Use principle of superposition and find mass of removed sphere

By the principle of superposition, the potential due to the remaining sphere with the cavity is the potential due to the entire original solid sphere minus the potential due to the removed spherical part. Since the sphere is uniform, its mass is directly proportional to its volume, meaning the removed sphere of radius $R/2$ has mass $M' = M/8$ .

Step 2: Calculate potential at cavity center due to original full sphere

The centre of the cavity is at a distance $r = R/2$ from the centre of the full sphere. The gravitational potential inside a solid uniform sphere at distance $r \le R$ is given by $V(r) = -\frac{GM}{2R^3}(3R^2 - r^2)$ . Substituting $r = R/2$ gives $3R^2 - R^2/4 = 11R^2/4$ , which evaluates to $-11GM/(8R)$ .

Step 3: Calculate potential at cavity center due to removed sphere

The point of interest is the exact centre of the removed sphere. The gravitational potential at the centre of a solid uniform sphere of mass $M'$ and radius $R' = R/2$ is given by $-\frac{3GM'}{2R'}$ . Substituting $M' = M/8$ and $R' = R/2$ , we obtain $-\frac{3G(M/8)}{R} = -\frac{3GM}{8R}$ .

Step 4: Subtract the two potentials to find net potential

Subtracting the contribution of the removed sphere from that of the entire sphere yields $V = -\frac{11GM}{8R} + \frac{3GM}{8R} = -\frac{8GM}{8R} = -\frac{GM}{R}$ . This corresponds to option (2).

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