Two point charges $\mathrm{q}_1$ and $\mathrm{q}_2$ are '$l$' distance apart. If one of the charges is doubled and distance between them is halved. The magnitude of force becomes n times, where n is

Let the initial charges be $q_1$ and $q_2$, separated by a distance $l$. The initial force magnitude is $F_1 = k\frac{|q_1 q_2|}{l^2}$.

When one charge is doubled ($q_1' = 2q_1$) and the distance halved ($l' = l/2$), the new force becomes:

$F_2 = k\frac{|q_1' q_2'|}{(l')^2} = k\frac{|(2q_1) q_2|}{(l/2)^2}$

Simplifying:

$F_2 = k\frac{2|q_1 q_2|}{l^2/4} = k\frac{2|q_1 q_2|}{l^2} \times 4 = 8\left(k\frac{|q_1 q_2|}{l^2}\right) = 8F_1$

The final force magnitude is 8 times the original force. The correct option is C.