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JEE MAINS 2024
31.01.24 S1
Question
Four identical particles of mass $m$ are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is $\left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^2}{\mathrm{~L}^2}$, the length of the sides of the square
Select the correct option:
A
L/2
B
4L
C
3L
D
2L
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. $$ \begin{aligned} & \mathrm{F}_{\mathrm{net}}=\sqrt{2} \mathrm{~F}+\mathrm{F}^{\prime} \\ & \mathrm{F}=\frac{\mathrm{Gm}^2}{\mathrm{a}^2} \text { and } \mathrm{F}^{\prime}=\frac{\mathrm{Gm}^2}{(\sqrt{2} \mathrm{a})^2} \\ & \mathrm{~F}_{\mathrm{net}}=\sqrt{2} \frac{\mathrm{Gm}^2}{\mathrm{a}^2}+\frac{\mathrm{Gm}^2}{2 \mathrm{a}^2} \\ & \left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^2}{\mathrm{~L}^2}=\frac{\mathrm{Gm}^2}{\mathrm{a}^2}\left(\frac{2 \sqrt{2}+1}{2}\right) \\ & \mathrm{a}=4 \mathrm{~L} \end{aligned} $$
Question Tags
JEE Main
Physics
Easy
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