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 = ∞, the potential at the center of the cavity thus formed is: ( G = gravitational constant) (a) (-2GM)/(3R) (b) -2GM/R (c) -GM/2R (d) -GM/R

Answer: (d). R/2OPV(P) = V(full M)− V(M/8)radius R, cavity radius R/2 Treat the body as a full sphere M plus a sphere of mass -M/8 (radius R/2) at the cavity. Full sphere at r = R/2 (inside): V₁ = -(GM)/(2R³)(3R² - R²/4) = -(11GM)/(8R) Removed sphere at its own centre: V₂ = -(3G(M/8))/(2(R/2)) = -3GM/8R V = V₁ - V₂ = -(11GM)/(8R) + 3GM/8R = -GM/R So the answer is (d) -GM/R.

From a solid sphere of mass M and radius R, a spherical portion of… | JEE Main 2015

Q27JEE Main 2015Difficult

Gravitational Potential & Energy

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 center of the cavity thus formed is:

( $G = \text{gravitational constant}$ )

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Answer: (d)

Treat the body as a full sphere $M$ plus a sphere of mass $-\frac{M}{8}$ (radius $\frac{R}{2}$ ) at the cavity.
Full sphere at $r = \frac{R}{2}$ (inside): $V_1 = -\frac{GM}{2R^3}\left(3R^2 - \frac{R^2}{4}\right) = -\frac{11GM}{8R}$
Removed sphere at its own centre: $V_2 = -\frac{3G(M/8)}{2(R/2)} = -\frac{3GM}{8R}$
$V = V_1 - V_2 = -\frac{11GM}{8R} + \frac{3GM}{8R} = -\frac{GM}{R}$
So the answer is (d) $-\frac{GM}{R}$ .

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