The work done in breaking a drop of liquid of radius $R$ (Surface tension $T$ ) into 64 equal drops is
The work done in breaking a drop of liquid of radius $R$ (Surface tension $T$ ) into 64 equal drops is
$4 \pi R^2 T$
$\frac{\pi R^2 T}{64}$
$\frac{12 \pi T}{R^2}$
$12 \pi R^2 T$
Solution
Radius of big drop $=R$
Number of small drops, $n=64$
Let radius of small drop $=r$
In the process of breaking the drop, the volume of liquid drops remains constant.
$V_i=V_f$
$\begin{aligned}
\Rightarrow \quad \frac{4}{3} \pi R^3 & =n\left(\frac{4}{3} \pi r^3\right) \\
R^3 & =64 r^3 \Rightarrow R=4 r \\
r & =\frac{R}{4}
\end{aligned}$
Now, work done in breaking a big drop in 64 small drops.
$\begin{aligned}
W & =\text { final energy }- \text { initial energy } \\
W & =T\left(n 4 \pi r^2\right)-T\left(4 \pi R^2\right)=4 \pi T\left[64\left(\frac{R}{4}\right)^2-R^2\right] \\
& =4 \pi T\left[4 R^2-R^2\right]=12 \pi R^2 T
\end{aligned}$