Question

A water drop of diameter 2 $cm$ is broken into $64$ equal droplets. The surface tension of water is $0.075\, N / m$. In this process the gain in surface energy will be :

• A. $2.8 \times 10^{-4} J$
• B. $1.5 \times 10^{-3} J$
• C. $1.9 \times 10^{-4} J$
• D. $9.4 \times 10^{-5} J$

Answer 1

Answer: (A)

$ d =2\, cm $;
$ r =1\, cm $;
$ T =0.075 $
$\Delta SE = T \Delta A $
$=0.075\left( A _{ f }- A _{1}\right) $
$ A _{ i }=4 \pi r ^{2} $
$ A _{ f }=4 \pi r _{0}^{2} \times 64$
By volume conservation
$\frac{4}{3} \pi r^{3}=64 \cdot \frac{4}{3} \pi r_{0}^{3}$
$r _{0}=\frac{ r }{4}$
$A _{ f }=4 \pi\left(\frac{ r }{4}\right)^{2} \cdot 64=16 \pi r ^{2}$
$\Delta SE =0.075\left(16 \pi r ^{2}-4 \pi r ^{2}\right)$
$=0.075\left(12 \pi(0.01)^{2}\right)$
$=2.8 \times 10^{-4} J$