Question

1000 small water drops of water, each of radius ${10}^{-7}m$ are falling down, coalesce to form a bigger drop. Find out the free energy. Surface tension of water is $7\times {10}^{-2}N/m$.

Answer 1

Area of big drop = 1000 $\times$ area of small drop

or $\frac{4}{3}\pi R^3=1000\times \frac{4}{3}\pi r^3$

or $R=10r=10\times {10}^{-7}$

or $R={10}^{-6}m$

$\therefore$ Reduction in area :

$\Delta A=A_1-A_2=1000\times 4\pi r^2-4\pi R^2$

$=4\pi (1000\times r^2-R^2)$

$=4\pi (1000\times {10}^{-14}-{10}^{-12})$

$=4\pi \times {10}^{-12}(10-1)$

or $\Delta A=4\times 3.14\times {10}^{-12}\times 9=36\times 3.14\times {10}^{-12}$

or $\Delta A=113.04\times {10}^{-12}{m}^2$

$\therefore$ Free energy

$\Delta E=T\Delta A=7\times {10}^{-2}\times 113.04\times {10}^{-12}$

$=7\times 1.13\times {10}^{-2}$

$=7.91\times {10}^{-2}J$