Question

Work done in splitting a drop of water of 1 mm radius into ${10}^6$ droplets is (Surface tension of water = 72 x ${10}^{-3}J/m^2$)

• A. 9.58 x ${10}^{-5}J$
• B. 8.95 x ${10}^{-5}J$
• C. 5.89 x ${10}^{-5}J$
• D. 5.98 x ${10}^{-6}J$

Answer 1

The correct option is B 8.95 x ${10}^{-5}J$

let R the radius of the big drop R= ${10}^{-3}$m

r is the radius of small drops

since volume remains constant $⟹\frac{4}{3}\pi R^3$ = n($\frac{4}{3}\pi r^2$)

$⟹$ n = $(\frac{R}{r}{)}^3$

intial area of the big drop = $4\pi R^2$

final drops (small) area = $n\times 4\pi r^2$

change in area = $4\pi (n{r}^2-R^2)$

work done W =$4\pi R^{3}T(\frac{1}{r}-\frac{1}{R})$

=$4\pi R^{2}T[n^{\frac{1}{3}}-1]$

=$4\pi \times ({10}^{-3}{)}^2\times 72\times {10}^{-3}\times [{10}^{\frac{6}{3}}-1]$

= $8.95\times {10}^{-5}J$