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JEE Main 2015 · 4 Apr · Shift 1, Physics question 7

From a solid sphere of mass M and radius R, a spherical portion of radius R2\dfrac R2 is removed, as shown in the figure. Taking gravitational potential V = 0 at r = ∞, the potential at the centre of the cavity thus formed is :

(G = gravitational constant)

A shaded sphere with a spherical cavity of radius R/2 touching its surface, its centre midway between the sphere's centre and surface.
Choose one option
Answer and worked solution

Official answer: B

Superpose the full sphere and a sphere of mass −M8-\frac M8 filling the cavity. At the cavity's centre (r=R2r=\frac R2):
Vfull=−GM2R3(3R2−R24)=−11GM8R.V_{full}=-\frac{GM}{2R^3}\left(3R^2-\frac{R^2}4\right)=-\frac{11GM}{8R}.
The removed sphere at its own centre gives −3G(M/8)2(R/2)=−3GM8R-\dfrac{3G(M/8)}{2(R/2)}=-\dfrac{3GM}{8R}. So
V=−11GM8R+3GM8R=−GMR.V=-\frac{11GM}{8R}+\frac{3GM}{8R}=-\frac{GM}R.

Topics:Gravitational field and potential
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