JEE Main 2013 — Mechanical Properties of Fluids Question with Solution

Let the radius of the drop at time $t=t$ be $r$ and at an instant $t=t+\mathrm{d}t$ be $r-\mathrm{d}r$

As surface area is given by, $A=4\pi r^2$,

$\therefore$ decrease in surface area is,

$\mathrm{d}A=4\pi \left(2r\mathrm{d}r\right)$

$\therefore$ Decrease in surface energy during time $\mathrm{d}t$ is,

$\mathrm{d}U=T\mathrm{d}A=T·8\pi r\mathrm{d}r ...\left(1\right)$

Decrease in volume during time $\mathrm{d}t$ is,

$\mathrm{d}V=4\pi r^2\mathrm{d}r$

$\therefore$ Decrease in mass during time $\mathrm{d}t$ is,

$\mathrm{d}m=\rho \mathrm{d}V=4\pi \rho r^2\mathrm{d}r$

$\therefore$ Heat required in vaporisation is,

$\mathrm{d}Q=L\mathrm{d}m=4\pi \rho r^2\mathrm{d}rL ...\left(2\right)$

Apply conservation of energy,

decrease in surface energy$=$heat required in vaporisation.

$\Rightarrow \mathrm{d}U=\mathrm{d}Q$

$\Rightarrow 8T\pi r\mathrm{d}r=4\pi \rho r^2\mathrm{d}rL$

$\Rightarrow 2T=\rho rL$

$\Rightarrow r=\frac{2T}{\rho L}$

Hence, the minimum radius is $\frac{2T}{\rho L}$