Practice portal › Gravitation › Newton's Law of Gravitation

Net gravitational force at the centre of a square is found to be F₁ when four particles of masses M, 2M, 3M and 4M are placed at the four corners of the square as shown in the 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

Asked in JEE Main 22nd Jan 1st Shift 2026 · Force due to a system of masses

• (1) 1
• (2) 2√5
• (3) 3
• (4) 2

Step-by-step solution

Why the other options are wrong

• (1) Comes from assuming that merely swapping two corner masses cannot change the resultant, so F₁=F₂.
• (2) Comes from rationalising (2√2)/(√10) into (2√5)/5 and then reading the numerator straight off as α, as if the denominator 5 were √5. The correct match needs (2√5)/5=2/(√5), so α=2.
• (3) Comes from keeping only one diagonal after the swap, where the mass difference is 4M-M=3M, and calling that the answer.

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.