JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Gravitation

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If V is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:-

(A) $\frac{3 V}{2}$ ✓ Correct
(B)V
(C) $\frac{4}{3} V$
(D) $\frac{V}{2}$

Answer: (A)

Solution

For a uniform sphere, the gravitational potential at distance r from center is

V (r) = \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})

. At the surface (r = R):

V_{s} = \frac{GM}{R}

. At the center (r = 0):

V_{c} = \frac{3 GM}{2 R} = \frac{3}{2} V_{s}

. Therefore, the potential at the center is

\frac{3 V}{2}

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option

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