JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Gravitation

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Three masses 200 kg, 300 kg and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process ______ J.
(Gravitational constant

G = 6.7 \times 1 0^{- 11} N m^{2} / kg^{2}

)

(A) $9.86 \times 1 0^{- 6}$
(B) $2.85 \times 1 0^{- 7}$
(C) $1.74 \times 1 0^{- 7}$ ✓ Correct
(D) $4.77 \times 1 0^{- 7}$

Answer: (C)

Solution

The work done equals the change in gravitational potential energy of the three pairs. For an equilateral triangle of side

d

U = - \frac{G}{d} (m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1})

, where

m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = 200 \cdot 300 + 300 \cdot 400 + 400 \cdot 200 = 2.6 \times 1 0^{5} kg^{2}

. Then

W = U_{f} - U_{i} = - G (2.6 \times 1 0^{5}) (\frac{1}{25} - \frac{1}{20}) = G (2.6 \times 1 0^{5}) (0.01)

. With

G = 6.7 \times 1 0^{- 11}

W = 6.7 \times 1 0^{- 11} \times 2600 = 1.74 \times 1 0^{- 7} J

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Gravitation

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