JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Gravitation

Four identical particles of mass

m

are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is

(\frac{2 2 + 1}{32}) \frac{Gm ^{2}}{L ^{2}}

, the length of the sides of the square is

Answer: (B)

Solution

Let the side of the square be

a

. The particle at one corner is attracted by

\frac{G m ^{2}}{a ^{2}}

from each of the two adjacent corners (at right angles, resultant

2 \frac{G m ^{2}}{a ^{2}}

along the diagonal), and by

\frac{G m ^{2}}{( a 2 ) ^{2}} = \frac{G m ^{2}}{2 a ^{2}}

from the diagonally opposite corner, acting along the same diagonal.

Net force:

F_{n e t} = 2 \frac{G m ^{2}}{a ^{2}} + \frac{G m ^{2}}{2 a ^{2}} = (\frac{2 2 + 1}{2}) \frac{G m ^{2}}{a ^{2}}

Given

F_{n e t} = (\frac{2 2 + 1}{32}) \frac{G m ^{2}}{L ^{2}}

, equating:

\frac{1}{2 a ^{2}} = \frac{1}{32 L ^{2}} ⟹ a^{2} = 16 L^{2} ⟹ a = 4 L

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option