Question

A large number of water drops, each of radius $r$, combine to form a drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$, the rise in heat energy per unit volume will be :

• A. $\frac{2T}{rJ}$
• B. $\frac{4T}{rJ}$
• C. $\frac{2T}{J}(\frac{1}{r}-\frac{1}{R})$
• D. $\frac{3T}{J}(\frac{1}{r}-\frac{1}{R})$

Answer 1

The correct option is D $\frac{3T}{J}(\frac{1}{r}-\frac{1}{R})$

$R$ is the radius of bigger drop.
$r$ is the radius of smaller drops.

Water drops are combined to form a bigger drop.

So,
Volume of $n$ smaller drops $=$ volume of a bigger drop

$\Rightarrow n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

$\Rightarrow n={\left(\frac{R}{r}\right)}^3$

Now,
$\Delta$ $U=T\times \text{change in surface area}$

$\Rightarrow \Delta U=T(n\times 4\pi r^2-4\pi R^2)$

$\Rightarrow \Delta U=4\pi T[{\left(\frac{R}{r}\right)}^3{r}^2-R^2]=4\pi T(\frac{R^3}{r}-R^2)$

$\Rightarrow H=\frac{\Delta U}{J}=\frac{4\pi T(\frac{R^3}{r}-R^2)}{J}$

Further,

$\frac{H}{V}=\frac{4\pi T(\frac{R^3}{r}-R^2)}{J\times \frac{4}{3}\pi R^3}=\frac{3T}{J}(\frac{1}{r}-\frac{1}{R})$