A particle of mass $m_{1}$ is kept at $x = 0$ and another particle of mass $m_{2}$ at $x = d$ . When a third particle is kept at $x = d / 4$ , it experiences no net gravitational force due to the two particles. Find $\frac{m_{2}}{m_{1}}$ .

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Solution

The correct option is C $9$

Suppose, the third particle has mass $m$ .

From equilibrium of forces,

$\frac{G m_{1} m}{(d / 4)^{2}} = \frac{G m_{2} m}{(d - d / 4)^{2}}$

$\Rightarrow \frac{m_{2}}{m_{1}} = \frac{(3 d / 4)^{2}}{(d / 4)^{2}} = 9$

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A particle of mass m1 is kept at x=0 and another particle of mass m2 at x=d. When a third particle is kept at x=d/4, it experiences no net gravitational force due to the two particles. Find m2m1.

The correct option is C 9 Suppose, the third particle has mass m. From equilibrium of forces, Gm1m(d/4)2=Gm2m(d−d/4)2 ⇒m2m1=(3d/4)2(d/4)2=9

A particle of mass $m_{1}$ is kept at $x = 0$ and another particle of mass $m_{2}$ at $x = d$ . When a third particle is kept at $x = d / 4$ , it experiences no net gravitational force due to the two particles. Find $\frac{m_{2}}{m_{1}}$ .

The correct option is C $9$

Suppose, the third particle has mass $m$ .

From equilibrium of forces,

$\frac{G m_{1} m}{(d / 4)^{2}} = \frac{G m_{2} m}{(d - d / 4)^{2}}$

$\Rightarrow \frac{m_{2}}{m_{1}} = \frac{(3 d / 4)^{2}}{(d / 4)^{2}} = 9$