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Two bodies P and Q having masses $2 kg$ and $4 kg$ respectively are separated by $2 m$ . Which of the following should be the distance from P, at which the mass R weighing $1 kg$ is kept,so that gravitational force on R due to P and Q is zero? (Gravitational constant $G = 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ )

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0.82 m

0.082 m

8.2 m

2 m

answer is A.

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Detailed Solution

The mass R which is of

1 kg

weight need to be kept at a distance of

0.82 m

from P so that the gravitational force on R from P and Q will be zero.
Let us consider that the mass R is kept at a point S at a distance of

x

from P.
As per Newton’s law of gravitation we know that

F = \frac{G m_{1} m_{2}}{d^{2}} . . . . . . . . . . (a)

where

G

is the gravitational constant (

6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}

)

m_{1}

is the mass of the first object.

m_{2}

is the mass of the second object.

d

is the distance between them.
We can refer to the figure below.
Screenshot_2022-08-25_at_07.10.17-removebg-preview

Considering P and R
mass of P,

m_{1} = 2 kg

mass of R,

m_{2} = 1 kg

distance between P and R,

d = x m

From (a) the gravitational force

F

on R due to P is given by

F = \frac{G m_{1} m_{2}}{x^{2}}

On substituting the values we get

F = \frac{G \times 2 \times 1}{x^{2}}

along SA
Here

G

is the Gravitational constant
Similarly we can write the gravitational force

F_{1}

on R due to Q and it is given by

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

mass of Q,

m_{3} = 4 kg

mass of R,

m_{2} = 1 kg

distance between Q and R,

d = (2 - x) m

On substituting the values we get

F_{1} = \frac{G \times 4 \times 1}{{(2 - x)}^{2}}

along SB
We are asked to find the distance of R from P such that the force on R due to P and Q is zero.
If the force on R due to P and Q is zero then

F = F_{1}

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

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

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

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

\Rightarrow {(2^{2} - (2 \times 2 \times x) + x}^{2}) = \frac{{4 x}^{2}}{2}

{\Rightarrow (4 - 4 x + x}^{2}) = {2 x}^{2}

{\Rightarrow (4 - 4 x + x}^{2}) - {2 x}^{2} =

\Rightarrow 4 - 4 x - x^{2} =

\Rightarrow

- x^{2} - 4 x + 4 = 0

Solving this equation we get

x = - 2 \mp 2 \sqrt{2}

As we need to find the distance, consider only the positive value so that

x = 0.82 m

.
So the distance from P to R is

0.82 m

such that the force on R due to P and Q is zero.

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