Question

'$n$' droplets of equal size of radius $r$ coalesce to form a bigger drop of radius $R$. The energy liberated is equal to _________.
($T=$ Surface tension of water)

• A. $4\pi R^{2}T[n^{1/3}-1]$
• B. $4\pi r^{2}T[n^{1/3}-1]$
• C. $4\pi R^{2}T[n^{2/3}-1]$
• D. $4\pi r^{2}T[n^{2/3}-1]$

Answer 1

The correct option is A $4\pi R^{2}T[n^{1/3}-1]$

As the mass of the water is conserved, thus mass of bigger drop is equal to the mass of $n$ droplets.

$\therefore$ $\rho \left(\frac{4\pi }{3}{R}^3\right)=\rho \left(n\frac{4\pi }{3}{r}^3\right)$ where $\rho$ is the density of water

$⟹$ $r=n^{-1/3}R$ .......(1)

Surface area of bigger drop $A=4\pi R^2$

Total Surface area of $n$ droplets $A^{\prime }=n\left(4\pi r^2\right)$

$\therefore$ $A^{\prime }=n4\pi \left(n^{-2/3}{R}^2\right)=4\pi R^2{n}^{1/3}$

Change in surface area $\Delta A=A^{\prime }-A=4\pi R^2[n^{1/3}-1]$

$\therefore$ Energy liberated $E=T\Delta A=4\pi R^{2}T[n^{1/3}-1]$