According to the question, the arrangement of the masses is as shown below,

KCET 2021 Physics - Gravitation Question 12 English Explanation

Gravitational force between two masses is given as

$$F=\frac{G m_1 m_2}{r^2}$$

where, $$G=$$ gravitational constant and $$r=$$ distance between them.

From the given values, we can say that force between masses at $$A$$ and $$G$$.

$$=$$ Force between masses at $$B$$ and $$G$$.

$$\begin{array}{ll} \Rightarrow & F_A=\frac{G m_A m_G}{A G^2} \text { and } F_B=\frac{G m_B m_G}{B G^2} \\ \text { or } & F_A=F_B=\frac{G \times 8 \times 2}{1^2}=16 G \quad \text{.... (i)} \end{array}$$

[$$\because$$ Given, $$m_1=m_A=m_B=8 \mathrm{~kg}, m_2=m_G=2 \mathrm{~kg}$$, $$A G=B G=r=1 \mathrm{~m}$$]

From the figure, resultant of $$F_A$$ and $$F_B$$ is given as

$$\begin{aligned} F_{A B} & =\sqrt{F_A^2+F_B^2+2 F_A F_B \cos \theta} \\ & =\sqrt{2 F_A^2+2 F_A^2 \cos 120^{\circ}} \\ & =\sqrt{F_A^2}=\sqrt{(16 G)^2}=16 G \end{aligned}$$

[from Eq. (i)]

$$\Rightarrow \quad F_{A B}=F_A=F_B=16 G$$

For resultant force on $$2 \mathbf{k g}$$ body to be zero, $$4 \mathrm{~kg}$$ body should be placed at a certain distance from $$G$$ such that

$$\mathbf{F}_{C G}=-\mathbf{F}_{A E}$$

$$\begin{aligned} & \Rightarrow \quad\left|\mathbf{F}_{C G}\right|=\left|\mathbf{F}_{A B}\right| \\ & \text { or } \frac{G \times m_C m_G}{x^2}=G(16) \\ \end{aligned}$$

where, $$x=$$ distance between $$4 \mathrm{~kg}$$ and $$2 \mathrm{~kg}$$ body.

Here, $$m_C=4 \mathrm{~kg}$$

$$\begin{array}{rlrl} \Rightarrow & \frac{G \times 4 \times 2}{x^2} =G(16) \\ \Rightarrow & x^2 =\frac{1}{2} \\ \Rightarrow & x =\frac{1}{\sqrt{2}} \mathrm{~m} \end{array}$$

This means, $$4 \mathrm{~kg}$$ body should be placed at point $$P$$ on line $$C G$$ such that $$P G=\frac{1}{\sqrt{2}} \mathrm{~m}$$.

Gravitation

Continue this paper

Similar Gravitation PYQs

More practice from this question