Question

If ‘n’ drops of a liquid each with surface energy E join to form a single drop, in this process:

• A. Some energy is absorbed
• B. Energy absorbed is $E(n-n^{2/3})$
• C. Energy released is $E(n-n^{2/3})$
• D. Energy released is $E(n^{2/3}-1)$

Answer 1

The correct option is C Energy released is $E(n-n^{2/3})$

$U_i=nE$ (Let radius be r)

$U_f=E^{\prime }$ (Let radius be R)
$n\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

$n{r}^3=R^3$

New surface area $=4\pi R^2$

$=4\pi n^{2/3}{r}^2$

New surface Energy $=n^{2/3}E$

$\Delta U=nE-n^{2/3}E$

$=(n-n^{2/3})E$
$U_i>U_f$ i.e. energy is released