JEE Main 2026 · 22 January · morning · Single correct · +4 / −1
Net gravitational force at the center of a square is found to be F₁ when four particles having mass M,2M,3M and 4M are placed at the four corners of the square as shown in figure and it is F₂ when the positions of 3M and 4M are interchanged. The ratio F₁/F₂ is α/√5. The value of α is ____ .
- (a) 2
- (b) 2√5
- (c) 1
- (d) 3
Answer and solution
Answer (a) 2
Solution
Physics sir · 3 steps
Pair each mass with the one across the diagonal: opposite pulls cancel, and only the leftover mass matters.
- Same distance. Every corner is r = a/√2 from the centre. So a mass kM pulls a test mass m at the centre with kF₀, where F₀ = GMm/r², directed towards that corner.
- Diagonal pairs. Masses at opposite corners pull in exactly opposite directions, so each diagonal gives a net pull of (difference of masses)× F₀. The two diagonals of a square are perpendicular, so the two net pulls add like the sides of a right triangle.
Case Diagonal 1 (bottom-left to top-right) Diagonal 2 (bottom-right to top-left) Net force F₁ (as drawn) M vs 3M: 2F₀ 2M vs 4M: 2F₀ √(2²+2²) F₀ = 2√2 F₀ F₂ (3M, 4M swapped) M vs 4M: 3F₀ 2M vs 3M: F₀ √(3²+1²) F₀ = √10 F₀ F₁/F₂ = 2√2/√10 = 2/√5
- So. Comparing with α/√5 gives α = 2, option (a).
The idea to keep. For masses at the corners of a square, cancel along each diagonal first. The two diagonal leftovers are perpendicular, so combine them with Pythagoras, not by adding.
Answer from the NTA final answer key. Solution by PrepWiser.