JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

Net gravitational force at the centre 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 corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F _{1}}{F _{2}}$ is $\frac{α}{5}$ . The value of $α$ is _____.

A
2
B
3
C
1
D
$25$

Solution

Place a test mass $m_{0}$ at the centre. Each corner contributes a force of magnitude $\frac{G \cdot ( corner mass ) \cdot m _{0}}{r ^{2}}$ where $r$ is the half-diagonal.

For the initial configuration (4M, 3M on top; M, 2M on bottom), diagonally opposite pairs partially cancel: 4M (top-left) opposite to 2M (bottom-right) leaves a net $2 M$ pull along that diagonal; 3M (top-right) opposite to M (bottom-left) leaves a net $2 M$ pull along the other diagonal. The two equal $\frac{2 GM m _{0}}{r ^{2}}$ contributions are perpendicular, summing to $F_{1} = 22 \frac{GM m _{0}}{r ^{2}}$ .

After swapping 3M and 4M: diagonal 1 has 3M opposite 2M (net $M$ ); diagonal 2 has 4M opposite M (net $3 M$ ). These two perpendicular contributions are $\frac{GM m _{0}}{r ^{2}}$ and $\frac{3 GM m _{0}}{r ^{2}}$ , giving $F_{2} = 1^{2} + 3^{2} \frac{GM m _{0}}{r ^{2}} = 10 \frac{GM m _{0}}{r ^{2}}$ .

Ratio: $\frac{F _{1}}{F _{2}} = \frac{2 2}{10} = \frac{2 2}{2 5} = \frac{2}{5}$ . So $α = 2$ .

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