Question

The diameter of one drop of water is $0.2cm$. The work done in breaking one drop into $1000$ equal droplets will be :-
(surface tension of water $=7\times {10}^{-2}N/m$)

• A. $7.9\times {10}^{-6}J$
• B. $5.92\times {10}^{-6}J$
• C. $2.92\times {10}^{-6}J$
• D. $1.92\times {10}^{-6}J$

Answer 1

The correct option is A $7.9\times {10}^{-6}J$

Current radius of a drop = 0.1 cm

On breaking it into 1000 droplets, the volume would remain constant. Therefore, radius of each small drop,

$1.33\pi R^3=1000\times 1.33\pi r^3$

Or $r=R/10=0.01$ cm

Change in surface area

$\delta A=1000\times 4\pi r^2-4\pi R^2$

= $4\pi (10R^2-R^2)$

= $36\pi R^2$

Therefore, the change in surface energy,

$\delta U=T\delta A$

$\delta U=7\times {10}^{-2}\times 36\pi \times {10}^{-6}$

$\delta U=7.91\times {10}^{-6}J$