Net gravitational force at the center of a square is found to be F_1 when four particles having mass M, 2 M, 3 M and 4 M are placed at the four…
JEE Main 2026 — Physics Mechanics
2026mcqmedium
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 F2F1 is 5α. The value of α is ____.
Official previous-year question
Held on 22 Jan 2026 · Verified 6 Jul 2026.
Options
A
25
B
3
C
1
D
2
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Solution
Let k=r22G where r=a/2.
Net field at center has components along x and y.
Config 1 (original):
gx=k(−M+2M+3M−4M)=0,
gy=k(−M−2M+3M+4M)=4kM.
F1=4kM.
Config 2 (3M and 4M swapped):
gx=k(−M+2M+4M−3M)=2kM,
gy=k(−M−2M+4M+3M)=4kM.
F2=kM4+16=25kM.
F2F1=254=52⇒α=2.
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