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The gravitational force acting on a particle of mass 1g due to a similar
particle is equal to $6.67 \times 10^{- 17} N$ .Then the separation between particles is

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1 m

\frac{1}{100} m

100 m

1 cm

answer is B.

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Detailed Solution

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

= $6.67 \times 10^{- 17} = \frac{6.67 \times 10^{- 11} \times 10^{- 3} \times 10^{- 3}}{r^{2}}$

$\Rightarrow r^{2} = 1 m$

$\Rightarrow r = 1m$

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