Question

Two mercury droplets of radii $0.1cm\text{and}0.2cm$ collapse into one single drop. What amout of energy is released? The surface tension of mercury $T=435.5\times {10}^{-3}\frac{N}{m}.$

HInt : ''Coalescence of two drops''

Formula used :

$Workdone=T\times changeinsufacearea$

Answer 1

Given,

Radius of small drop $1,\left(r_1\right)=0.1cm$

Radius of small drop $2,\left(r_2\right)=0.2cm$

Radius of big drop $=r$

In the coalescence of drop volume remain conserved.

$V=V_1+V_2$

$\frac{4}{3}\pi r^3=\frac{4}{3}\pi r_1^3+\frac{4}{3}\pi r_2^3$

$r^3=r_1^3+r_2^3$

$r=({0.1}^3+{0.2}^2{)}^{\frac{1}{3}}$

$r=0.21cm$

As surface area is decreasing during coalescence of two small drops, hence energy will releae in this process.

$\Delta U=T\times changeinsufacearea$

$\Delta U=T\Delta A=T(4\pi r^2-(4\pi r_1^2+4\pi r_2^2\left)\right)$

$\Delta U==435.5\times {10}^{-3}\times 4\pi ({0.21}^2-({0.1}^2+{0.2}^2\left)\right){\times }^{-4}$

$\Delta U=-32.28\times {10}^{-7}J$

Final Answer : $-32\times {10}^{-7}J$