Question

If $T$ is the surface tension of a liquid, the energy needed to break a liquid drop of radius $R$ into 64 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 $=T$
Number of smaller drop, $n=64$
Let us suppose, radius of smaller drop is $r$.
As volume of liquid will be same.
Volume of larger drop $=$ Volume of $n$ smaller drops
$\Rightarrow \frac{4}{3}\pi R^3=n\times \frac{4}{3}\pi r^3$
$\Rightarrow R^3=64\times r^3$
$\Rightarrow r=\frac{R}{4}$ .........(1)
We know that, change in energy $=$ surface tension $\times$ change in surface area
$\Rightarrow \Delta E=T\Delta A$
$=T\times 4\pi [64\times (R/4{)}^2-\left(R{)}^2\right)]$
$[$ from (1) and surface area $=4\pi r^2]$
$=12\pi R^{2}T$