Question

What is the work done in breaking a drop of water of 4 mm diameter into 1000 drops of same size, surface tension of water is $72 \times 10^{-3} Nm ^{-1} ?$

• A. $1.125 \times 10^{-4} J$
• B. $1.125 \times 10^{-5} J$
• C. $3.25 \times 10^{-5} J$
• D. $3.25 \times 10^{-6} J$

Answer 1

Answer: (C)

Volume remains constant. Let $R$ and $r$ be the radius of bigger and smaller drops respectively.
$\frac{4}{3} \pi R ^{3}=10^{3} \times \frac{4}{3} \pi r ^{3} $
$R =10 \times r $
$\Rightarrow r =\frac{ R }{10}=\frac{2 \times 10^{-3}}{10}=2 \times 10^{-4} m$
Work done in breaking drop into 1000 drops.
$W = S \times\left(10^{3} \times 4 \pi r ^{2}-4 \pi R ^{2}\right)= S \times 4 \pi\left(10^{3} \times r ^{2}- R ^{2}\right)$
$ W=72 \times 10^{-3} \times 4 \times 3.14\left[10^{3} \times\left(2 \times 10^{-4}\right)^{2}-\left(2 \times 10^{-3}\right)^{2}\right]$
$W=0.904\left(10^{3} \times 4 \times 10^{-8}-4 \times 10^{-6}\right)$
$W=0.904\left(4 \times 10^{-5}-4 \times 10^{-6}\right)$
$W=0.904 \times 4 \times 10^{-5}(1-0.1)=3.25 \times 10^{-5} J$