Two identical conducting balls A and B have positive charges q_1...

According to coulomb's law, the force of repulsion between them is given by $F=\frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}$ When the charged spheres $A$ and $B$ are brought in contact, each sphere will attain equal charge $q^{\prime}$ $q^{\prime}=\frac{q_1+q_2}{2}$ Now, the force of repulsion between them at the same distance $r$ is F^{\prime}=\frac{q^{\prime} \times q^{\prime}}{4 \pi \varepsilon_0 r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(\frac{q_1+q_2}{2}\right)\left(\frac{q_1+q_2}{2}\right)}{r^2} =\frac{\left(\frac{q_1+q_2}{2}\right)^2}{4 \pi \varepsilon_0 r^2} \operatorname{As}\left(\frac{q_1+q_2}{2}\right)^2 \gt q_1 q_2 \therefore F^{\prime} \gt F