JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Gravitation

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Four spheres each of mass m form a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is :

(A) $- \frac{Gm}{d} [(4 + 2) m + 42 M]$ ✓ Correct
(B) $- \frac{Gm}{d} [(4 + 2) M + 42 m]$
(C) $- \frac{Gm}{d} [3 m^{2} + 42 M]$
(D) $- \frac{G m}{d} [6 m^{2} + 42 M]$

Answer: (A)

Solution

Total PE has three contributions: (1) Four pairs of adjacent spheres at distance

d

- \frac{G m ^{2}}{d} \times 4

. (2) Two pairs of diagonal spheres at distance

2 d

- \frac{G m ^{2}}{2 d} \times 2

. (3) Four pairs of central sphere M with corner spheres at distance

\frac{d}{2}

- \frac{GM m}{d / 2} \times 4 = - \frac{4 2 GM m}{d}

. Combining:

- \frac{G m}{d} [4 m + 2 m + 42 M] = - \frac{G m}{d} [(4 + 2) m + 42 M]

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Gravitation

Solution

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