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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 1st object.

m_{2}

is the mass of the 2nd 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

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}

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} = 0

\Rightarrow 4 - 4 x - x^{2} = 0

\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

R

0.82 m

such that the force on

R

due to

P

and

Q

is zero.

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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 =6.67×10-11N m 2kg-2)

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