Two identical charged spheres are suspended by strings of equal...

For first case:

Tcos\theta = mg$and </p><p>Tsin\theta = {F}_{e}$

Therefore, $tan\theta = \dfrac{{F}_{e}}{mg}$

$tan\theta = \dfrac{{F}_{e}}{{ρ}_{B}Vg} . . . . ( i )$

For second case:

$tan\theta = \dfrac{\dfrac{{F}_{e}}{k}}{{ρ}_{B}Vg - {ρ}_{L}Vg} . . . . . ( ii )$ 

From Eq. (i) & (ii), we get

${ρ}_{B}Vg = \left({ρ}_{B} - {ρ}_{L}\right)kVg$

$⇒ 1 . 4 = 0 . 7k$

$⇒ k = 2$