Question

Calculate the energy released when $1000$ small water drops each of same radius ${10}^{-7}\text{m}$ coalesce to form one large drop. The surface tension of water is $5\times {10}^{-2}\text{N/m}$.

• A. $5\pi \times {10}^{-12}\text{J}$
• B. $4.2\pi \times {10}^{-12}\text{J}$
• C. $3.6\pi \times {10}^{-12}\text{J}$
• D. $1.8\pi \times {10}^{-12}\text{J}$

Answer 1

The correct option is D $1.8\pi \times {10}^{-12}\text{J}$

Let $r$ be the radius of smaller drop and $R$ of the bigger one.
Equating the initial and final volumes, we have
$\frac{4}{3}\pi R^3=1000\times \frac{4}{3}\pi r^3$
$\Rightarrow R=10r=10\times {10}^{-7}\text{m}$
As we know,
Water drops have only one free surface.
$\therefore$ Net Change in surface area;$\Delta A=4\pi R^2-1000\times 4\pi r^2$
$=4\pi ({10}^{-6}{)}^2-1000\times 4\pi ({10}^{-7}{)}^2$
$=-36\pi \times {10}^{-12}{\text{m}}^2$
Here, negative sign implies that surface area is decreasing. Hence energy released in the process,
$W=T\times |\Delta A|=0.05\times 36\pi \times {10}^{-12}{\text{m}}^2=1.8\pi \times {10}^{-12}\text{J}$