Question

Energy released when $1000$ small water drops each of same radius ${10}^{-7}\text{m}$ coalesce to form one large drop
(Surface tension of water is $7\times {10}^{-2}\text{N/m}$)

• A. $2\pi \times {10}^{-12}\text{J}$
• B. $2.5\pi \times {10}^{-12}\text{J}$
• C. $1.5\pi \times {10}^{-12}\text{J}$
• D. $5\pi \times {10}^{-12}\text{J}$

Answer 1

The correct option is B $2.5\pi \times {10}^{-12}\text{J}$

Given,
Number of water drops, $n=1000$
Radius of water drop,$r={10}^{-7}\text{m}$
Total volume $=n\times \frac{4}{3}\pi r^3$
$=1000\times \frac{4}{3}\pi \times ({10}^{-7}{)}^3$ ... (1)
Volume of large drop if its radius is $R$
$=\frac{4}{3}\pi R^3$ ... (2)
On equating (1) and (2) as volume will remain same,
$\frac{4}{3}\pi R^3=1000\times \frac{4}{3}\pi \times ({10}^{-7}{)}^3$
$\Rightarrow R={10}^{-6}\text{m}$
We know, surface energy, $U=T\Delta A$
$U=T\Delta A=T(4\pi R^2-n\times 4\pi r^2)$
$U=7\times {10}^{-2}\times 4\times \pi \times \left[\right({10}^{-6}{)}^2-1000\times \left({10}^{-7}{)}^2\right]$
$\Rightarrow U=-2.5\pi \times {10}^{-12}\text{J}$
(-ve sign indicates release of energy)