JEE Main 2026 · 24 January · morning · Single correct · +4 / −1
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× 10⁻¹¹ N m²/kg² )
- (a) 4.77× 10⁻⁷
- (b) 1.74× 10⁻⁷
- (c) 9.86× 10⁻⁶
- (d) 2.85× 10⁻⁷
Answer and solution
Answer (b) 1.74× 10⁻⁷
Solution
Physics sir · 4 steps
Work done equals the change in potential energy, and only the side length changes, so take the difference.
- Potential energy. Each pair of masses at separation a has energy -(G mᵢ mⱼ)/a. On an equilateral triangle all three pairs are the same distance apart, so U = -G/a(m₁ m₂ + m₂ m₃ + m₃ m₁).
- Pair products. 200 × 300 + 300 × 400 + 400 × 200 = 60000 + 120000 + 80000 = 2.6 × 10⁵ kg².
W = U_f - Uᵢ = G (2.6 × 10⁵)(1/20 - 1/25)
- Put in numbers. The bracket is 0.05 - 0.04 = 0.01 m⁻¹. So W = 6.7 × 10⁻¹¹ × 2.6 × 10⁵ × 0.01 = 6.7 × 10⁻¹¹ × 2600.
- So. W = 1.74 × 10⁻⁷ J. It is positive because the masses are pulled further apart against gravity. The answer is (b).
The idea to keep. For masses on an equilateral triangle, U = -G/aΣ mᵢ mⱼ. Moving them to a new triangle costs exactly U_f - Uᵢ, and the centre does not matter.
Answer from the NTA final answer key. Solution by PrepWiser.