Question:
27 similar drops of mercury are maintained at 10
$\mathrm{V}$ each. All these spherical drops combine into
a single big drop. The potential energy of the
bigger drop is ............ times that of a smaller drop.
Solution:
$(27)\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi R^{3}$
$R=3 r$
Potential energy of smaller drop :
$\mathrm{U}_{1}=\frac{3}{5} \frac{\mathrm{kq}^{2}}{\mathrm{r}}$
Potential energy of bigger drop :
$U=\frac{3}{5} \frac{k Q^{2}}{R}$
$U=\frac{3}{5} \frac{k(27 q)^{2}}{R}$
$U=\frac{3}{5} k \frac{(27)(27) q^{2}}{3 r}$
$U=\frac{(27)(27)}{3}\left(\frac{3}{5} \frac{k^{2}}{r}\right)$
$U=243 U_{1}$
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