TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Given, four masses are kept at corner of square and side of squares be $a$,As we know that, gravitational force, $F=\;\frac{G M^{2}}{d^{2}}$ where, $G$ is gravitational constant and $d$ is the separation between two bodies. If we calculate net force on 2 , then Since, $F_{21}=F_{23}=$ force on 2 due to 1 and 3 will be equal. $\therefore \; F_{21}=F_{23}=\; \frac{G M^{2}}{a^{2}}$ $\; \text{ and } \; \; \; F_{24} = \; \frac{G M^{2}}{(a \sqrt{2})^{2}} = \; \frac{G M^{2}}{2 a^{2}}$ Now, as forces are vector quantity and angle between $F_{21}$ and $F_{23}$ is $90^{\circ}$. $\therefore \; \; F_{\;\text{net}\;} \; \; = F_{24} + \sqrt{F_{21}^{2}+F_{23}^{2}}$ $\; \; = \; \frac{G M^{2}}{2 a^{2}} + \sqrt{ \left( \; \frac{G M^{2}}{a^{2}} \right)^{2} + \left( \; \frac{G M^{2}}{a^{2}} \right)^{2} }$ $\; \; = \; \frac{G M^{2}}{2 a^{2}} + \sqrt{2 \left( \; \frac{G M^{2}}{a^{2}} \right)^{2}}$ $\; \; = \; \frac{G M^{2}}{2 a^{2}} + \; \frac{G M^{2}}{a^{2}} \sqrt{2}$ $\; \; = \; \frac{G M^{2}}{a^{2}} \left( \; \frac{1}{2} + \sqrt{2} \right)$ $\; \; = \; \frac{G M^{2}}{a^{2}} \left( \; \frac{1+2\sqrt{2}}{2} \right)$ According to the question, $F_{\;\text{net}\;} = \left( \; \frac{2\sqrt{2}+1}{32} \right) \; \frac{G M^{2}}{L^{2}}$ Therefore, $a^{2}=16 L^{2}$ $a=4L$