Question

Calculate the energy released when 8000 small water droplets, each of same radius $2\times {10}^{-7}\text{m}$ coalesce to form one large drop. The surface tension of water is $7.5\times {10}^{-2}\text{N/m}$

• A. $4.2\times {10}^{-10}\text{J}$
• B. $2.8\times {10}^{-10}\text{J}$
• C. zero
• D. $6.5\times {10}^{-10}\text{J}$

Answer 1

The correct option is B $2.8\times {10}^{-10}\text{J}$

Let $r$ be the radius of smaller drops and $R$ of the bigger one.
Equating the initial and final volume, we have
$\Rightarrow \frac{4}{3}\pi R^3=8000\times \left(\frac{4}{3}\pi r^3\right)$
$\Rightarrow R^3=8000r^3$
$\Rightarrow R=(8000{)}^{\frac{1}{3}}r$
$\Rightarrow R=20r=20\times 2\times {10}^{-7}=4\times {10}^{-6}\text{m}$

Increase in surface area $\Delta A=4\pi R^2-8000\left(4\pi r^2\right)$
$=4\pi \times 400r^2-8000\times 4\pi r^2$
$=-7600\times 4\pi \times r^2$
$=-7600\times 4\pi \times 4\times {10}^{-14}$
[-ve sign shows surface area is decreasing ]
Energy released in the process
$U=T|\Delta A|=7.5\times {10}^{-2}\times 7600\times 16\pi \times {10}^{-14}$
$=2.8\times {10}^{-10}\text{J}$