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Two bodies of masses $3 kg$ and $12 kg$ are placed at a distance of $12 m$ . The third body of mass of $0.5 kg$ is to be placed at such a point that the force acting on this body is zero. The position of that point from $3 kg$ the body is:

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5 m

3 m

4 m

2 m

answer is C.

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Detailed Solution

Two bodies of masses

3 kg

and

12 kg

are placed at a distance of

12 m

. The third body of mass of

0.5 kg

is to be placed at such a point that the force acting on this body is zero. The position of that point from

3 kg

the body is

4 m

.
Let, mass of first body,

m_{1} = 3 kg

Mass of second body,

m_{2} = 12 kg

Mass of third body,

m_{3} = 0.5 kg

Let the mass

m_{3} = 0.5 kg

be placed at a distance of ‘

x

’ from

m_{1}

and at a distance of

(12 - x)

from

m_{2}

.
Gravitational force between first body and third body

F_{31} = \frac{G m_{1} m_{3}}{x^{2}}

…………… (1)
Gravitational force between second body and third body

F_{32} = \frac{G m_{2} m_{3}}{{(12 - x)}^{2}}

……………… (2)
Since the resultant force on a third body is zero, therefore

F_{31} = F_{32}

\Rightarrow \frac{G m_{1} m_{3}}{x^{2}} = \frac{G m_{2} m_{3}}{{(12 - x)}^{2}}

\Rightarrow \frac{3}{x^{2}} = \frac{12}{(12 - x)^{2}}

\Rightarrow \frac{1}{x^{2}} = \frac{4}{(12 - x)^{2}}

\Rightarrow \frac{1}{x^{}} = \frac{2}{(12 - x)^{}}

\Rightarrow 12 - x = 2 x

\Rightarrow 3 x = 12

\Rightarrow x = 4 m

.
Hence, the third body is kept

4 m

from the mass of

3 kg

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