Explanation

(b) Original charges on spheres $A$ and $B$ be $q_{1}\ and\ q_{2}$ respectively.
Distance between the two spheres =$r$
Since, both the spheres are of same size, they will possess equal charges on being brought in contact.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ q_{1}^{'} = \frac{q_{1} + q_{2}}{2}$
Similarly $q_{2\ }^{'} = \frac{q_{1} + q_{2}}{2}$
Therefore, new force of repulsion between spheres $A\ \ and\ B$ is
$F^{'} = \frac{1}{4\pi\varepsilon_{0}}\frac{\left\lbrack \frac{q_{1} + q_{2}}{2} \right\rbrack\left\lbrack \frac{q_{1} + q_{2}}{2} \right\rbrack}{r^{2}}$
$\ \ \ \ \ \ \ \ \ \ = \frac{\left\lbrack \frac{q_{1} + q_{2}}{2} \right\rbrack^{2}}{4\pi\varepsilon_{0}r^{2}}$
As $\left\lbrack \frac{q_{1 + q_{2}}}{2} \right\rbrack^{2} > q_{1}q_{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ F^{'} > F$