Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be
$(a)$. 1/4 times
$(b)$. 4 times
$(c)$. 1/2 times
$(d)$. unchanged

We know that force of attraction between two objects

$F=G\frac{m_1m_2}{r^2}$

Here, $F\rightarrow$ gravitational force between the two objects

$m_1\rightarrow$mass of the first particle

$m_2\rightarrow$mass of the second particle

$r\rightarrow$ distance between the two particles

If the distance between the two-particle is unchanged and the

masses are doubled, then force of attraction $F'=G\frac{(2m_1)(2m_2)}{r^2}$

Or $F'=4\times(G\frac{m_1m_2}{r^2})$

Or $F'=4F$

So, keeping the distance between them unchanged,  the mass of each

of the two particles is doubled, the value of gravitational force between them will become 4 times.

Therefore, option $(b)$ is correct.