JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 1)
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 $$\left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^2}{L^2}$$, the length of the sides of the square is
4L
3L
2L
$$\frac{L}{2}$$
Explanation
_31st_January_Morning_Shift_en_1_1.png)
$$\begin{aligned} & F_{\text {net }}=\sqrt{2} F+F^{\prime} \\ & F=\frac{G m^2}{a^2} \text { and } F^{\prime}=\frac{G^2}{(\sqrt{2} \mathrm{a})^2} \\ & F_{\text {net }}=\sqrt{2} \frac{\mathrm{Gm}^2}{\mathrm{a}^2}+\frac{\mathrm{Gm}^2}{2 \mathrm{a}^2} \\ & \left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^2}{\mathrm{~L}^2}=\frac{\mathrm{Gm}^2}{\mathrm{a}^2}\left(\frac{2 \sqrt{2}+1}{2}\right) \\ & \mathrm{a}=4 \mathrm{~L} \end{aligned}$$
Comments (0)
