Two identical charged spheres are suspended by strings of equal lengths. The string make an angle of 37 °…

For first case:

$Tcos\theta =mg$ and 

$Tsin\theta =F_e$

Therefore, $tan\theta =\frac{F_e}{mg}$

$tan\theta =\frac{F_e}{{\rho }_{B}Vg}....\left(\mathrm{i}\right)$

For second case:

$tan\theta =\frac{\frac{F_e}{k}}{{\rho }_{B}Vg-{\rho }_{L}Vg}.....\left(\mathrm{ii}\right)$ 

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

${\rho }_{B}Vg=\left({\rho }_B-{\rho }_L\right)kVg$

$\Rightarrow 1.4 = 0.7 k$

$\Rightarrow k = 2$