Q72 / 90 · JEE Main 2023 (01 Feb Shift 1) · Physics

A mercury drop of radius $10^{- 3} m$ is broken into $125$ equal size droplets. Surface tension of mercury is $0.45 N m^{- 1}$ . The gain in surface energy is:

Solution

As the volume of the mercury will remain same, therefore

$\frac{4}{3} π R^{3} = 125 (\frac{4}{3} π r^{3}) \Rightarrow R = 5 r$

Now, increase in surface energy

$∆ U = 125 \times (S \times 4 π r^{2}) - (S \times 4 π R^{2})$

$= 0.45 \times 4 π \times {(10^{- 3})}^{2} [\frac{125}{25} - 1]$

$= 4 \times 0.45 \times 4 π \times 10^{- 6}$

$= 2.26 \times 10^{- 5} J$

Hence, the correct answer is option: 'A'.