Question

If $T$ is the surface tension of a liquid, then the energy needed to break the liquid drop of radius $R$ into $64$ identical drops is

• A. $6\pi R^{2}T$
• B. $2\pi R^{2}T$
• C. $12\pi R^{2}T$
• D. $8\pi R^{2}T$

Answer 1

The correct option is C $12\pi R^{2}T$

Given,
Radius of larger drop $=R$
Surface Tension of liquid $=T$
Number of identical drops made from larger drop, $n=64$
Let us suppose, radius of smaller drop is $r$.
As volume of liquid will remain same.
$\Rightarrow$ Volume of larger drop $=n\times$ volume of smaller drop
$\Rightarrow \frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$
$\Rightarrow r=\frac{R}{4}$
We know that,
Work done $=$ Surface Tension $\times$ change in surface area
$\Rightarrow W=T\times \Delta A$
$=T(n\times 4\pi r^2-4\pi R^2)$
$=T[64\times 4\pi {\left(\frac{R}{4}\right)}^2-4\pi R^2]$
$=12\pi R^{2}T$