Physics · Gravitation

JEE Main 2026 — 22 January, Morning Shift — Question 33

Net gravitational force at the centre of a square is found to be F1F_{1} when four particles having mass M,2M,3MM, 2 M, 3 M and 4M4 M are placed at the four corners of the square as shown in figure and it is F2F_{2} when the positions of 3M3 M and 4M4 M are interchanged. The ratio F1F2\frac{F_{1}}{F_{2}} is α5\frac{\alpha}{\sqrt{5}}. The value of α\alpha is ____\_\_\_\_ .

Question figure
  1. Option A:

    2

    Correct
  2. Option B:

    3

  3. Option C:

    1

  4. Option D:

    252 \sqrt{5}

Answer: A

Step-by-step solution

Initial configuration F′=10Gmm0r2⇒ F F′=22⋅110=25\mathrm{F}^{\prime}=\sqrt{10} \frac{\mathrm{Gmm}_{0}}{\mathrm{r}^{2}} \Rightarrow \frac{\mathrm{~F}}{\mathrm{~F}^{\prime}}=2 \sqrt{2} \cdot \frac{1}{\sqrt{10}}=\frac{2}{\sqrt{5}} ∴α=2\therefore \alpha=2

figure

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Gravitation
Topic
Newton's Law of Gravitation
Net gravitational force at the centre of a square is found to be F 1… | JEE Main 2026 PYQ with Solution · DhiX AI