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 $0.075N/m)$

• 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
$=4\pi R^2\left(n^{1/3}-1\right)\times 0.075$
$=6.75\pi \times 10^{-15}J$