Question

A mercury drop of radius $1\text{cm}$ is sprayed into ${10}^6$ droplets of equal size. The energy expended in this process is -

(surface tension of mercury is equal to $32\times {10}^{-2}\text{N/m}$)

• A. $0.08\pi \text{J}$
• B. $0.03\pi \text{J}$
• C. $0.01\pi \text{J}$
• D. $0.05\pi \text{J}$

Answer 1

The correct option is C $0.01\pi \text{J}$

Given,
Radius of larger mercury drop, $R=1\text{cm}={10}^{-2}\text{m}$
Surface Tension, $T=32\times {10}^{-2}\text{N/m}$
Number of smaller mercury drops, $n={10}^6$
Let us suppose, radius of smaller mercury drop is $r$.
As volume of mercury will be same,
Volume of larger drop $=$ Volume of $n$ smaller drops
$\Rightarrow \frac{4}{3}\pi R^3=n\times \frac{4}{3}\pi r^3$
$\Rightarrow ({10}^{-2}{)}^3={10}^6\times r^3$
$\Rightarrow r={10}^{-4}\text{m}$ .........(1)
We know that, change in energy $=$ surface tension $\times$ change in surface area
$\Rightarrow \Delta E=T\Delta A$
$=32\times {10}^{-2}\times 4\pi [{10}^6\times ({10}^{-4}{)}^2-\left({10}^{-2}{)}^2\right)]$
$[$ from (1) and surface area $=4\pi r^2]$
$\approx 32\times {10}^{-2}\times 4\pi \times {10}^{-2}$
$\approx 128\pi \times {10}^{-4}$
$\approx 0.01\pi \text{J}$
So, energy of $0.01\pi \text{J}$ has to be expended in the process.