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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.