Question

$1000$ spherical drops of water each $10^{-8}\, m$ in diameter coalesce to form one large spherical drop. The amount of energy liberated in this process (in joule) is (surface tension of water is $0.075 \,N / m )$

• A. $10.75 \pi \times 10^{-15}$
• B. $6.75 \pi \times 10^{-15}$
• C. $8.65 \pi \times 10^{-15}$
• D. $3.88 \pi \times 10^{-15}$

Answer 1

Answer: (B)

Volume of small spherical drops $=\frac{4}{3} \pi \,r^{3}$
Volume of larger spherical drops $=\frac{4}{3} \pi \,R^{3}$
$n$ small spherical drops combine to form large drops.
$\therefore n\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi R^{3}$
$\Rightarrow R=n^{1 / 3} r$
Here, $n=1000, r=\frac{1}{2} \times 10^{-8} \,m$
Amount of energy liberated
$=$ Change in area $\times$ surface tension
$=4 \pi R^{2}\left(n^{1 / 3}-1\right) \times 0.075 $
$=6.75 \pi \times 10^{-15}\, J$