Two metal spheres of radii 0.01 m and 0.02 m are given a charge of 15 mC and 45 mC, respectively.
They are then connected by a wire. The final charge on the first sphere is ... $\times {10}^{-3} \mathrm{C}$. [Karnataka CET 2013]

When two metal spheres are joined by a wire, charge flows from one at higher potential tv the other at lower potential, till they acquire the same potential.

$\therefore$                             V₁=V₂

                          $\frac{{\mathrm{q}}_1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_1}=\frac{{\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_2}\Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=\frac{15\mathrm{mC}}{45\mathrm{mC}}=\frac{1}{3}$

Final charge on first sphere,

                           ${\mathrm{q}}_1=\frac{1}{3}\left({\mathrm{q}}_1+{\mathrm{q}}_2\right)=\frac{1}{3}\times 60=20\mathrm{mC}$