Question

$1000$ droplets of water having $2mm$ diameter each coalesce to form a single drop. Given the surface tension water is $0.72N{m}^{-1}$. The energy loss in the process is

• A. $8.146\times {10}^{-4}J$
• B. $4.4\times {10}^{-4}J$
• C. $2.108\times {10}^{-5}J$
• D. $4.7\times {10}^{-1}J$

Answer 1

Answer: (A)

$8.146\times {10}^{-4}J$

Let the radius of the large drop formed be $r$.

Since the amount of water must be same before and after the coalesce,

$1000\times \frac{4}{3}\pi (1\times {10}^{-3}{)}^3=\frac{4}{3}\pi r^3$

$⟹r={10}^{-2}m$

Hence the loss in surface energy in the process=Total surface energy of small drops-Surface energy of larger drop

$=0.72\times \left[\right(1000\times 4\pi (1\times {10}^{-3}{)}^2)-4\pi (1\times {10}^{-2}{)}^2]J$

$=8.146\times {10}^{-4}J$