Question

Work done in splitting a drop of water of $1mm$ radius into $64$ droplets is (surface tension of water $72\times {10}^{-3}J/m^2$).

• A. $2.0\times {10}^{-6}J$
• B. $2.4\times {10}^{-6}J$
• C. $4\times {10}^{-6}J$
• D. $5.4\times {10}^{-6}J$

Answer 1

The correct option is B $2.4\times {10}^{-6}J$

Volume of bigger drop = 64 (volume of smaller drop)

Let R = Radius of bigger drop and r = radius of smaller drop

$\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$
$r=\frac{R}{4}=\frac{{10}^{-3}}{4}$
$W=T\times (64\times 4\pi r^2-4\pi R^2)$
$W=72\times {10}^{-3}\times 4\pi [64\times {\left(\frac{{10}^{-3}}{4}\right)}^2-({10}^{-3}{)}^2]$
$W=2.7\times {10}^{-6}J$