Gravitation Question 9
Question 9 - 2024 (31 Jan Shift 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 $\left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^{2}}{\mathrm{~L}^{2}}$, the length of the sides of the square is
(1) $\frac{L}{2}$
(2) $4 \mathrm{~L}$
(3) $3 \mathrm{~L}$
(4) $2 \mathrm{~L}$
Show Answer
Answer: (2)
Solution:
$F_{\text {net }}=\sqrt{2} F+F^{\prime}$
$F=\frac{G m^{2}}{a^{2}}$ and $F^{\prime}=\frac{G m^{2}}{(\sqrt{2} \mathrm{a})^{2}}$
$F_{\text {net }}=\sqrt{2} \frac{G m^{2}}{a^{2}}+\frac{G m^{2}}{2 a^{2}}$
$\left(\frac{2 \sqrt{2}+1}{32}\right) \frac{G m^{2}}{L^{2}}=\frac{G^{2}}{a^{2}}\left(\frac{2 \sqrt{2}+1}{2}\right)$
$\mathrm{a}=4 \mathrm{~L}$
