A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne $/\mathrm{cm}$, then the gain in surface energy upto first decimal place will be : [Given $\pi=3.14$]

(C) $7.5 \times 10^{-4} \mathrm{~J}$

(A) $8.5 \times 10^{-4} \mathrm{~J}$

(B) $8.2 \times 10^{-4} \mathrm{~J}$

(D) $5.3 \times 10^{-4} \mathrm{~J}$

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Fluid Mechanics

A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne

/ cm

, then the gain in surface energy upto first decimal place will be :
[Given

π = 3.14

]

Answer: (C)

By volume conservation:

\frac{4}{3} π R^{3} = 729 \times \frac{4}{3} π r^{3}

, giving

r = \frac{R}{9} = \frac{1}{9}

cm.

Initial surface energy:

U_{i} = T \times 4 π R^{2} = 75 \times 1 0^{- 3} \times 4 π \times (1 0^{- 2})^{2} = 942 \times 1 0^{- 7}

J

Final surface energy:

U_{f} = 729 \times T \times 4 π r^{2} = 729 \times 75 \times 1 0^{- 3} \times 4 π \times \frac{( 1 0 ^{- 2} ) ^{2}}{81} = 9 \times 942 \times 1 0^{- 7}

J

Gain in surface energy:

Δ U = (9 - 1) \times 942 \times 1 0^{- 7} = 7536 \times 1 0^{- 7} = 7.5 \times 1 0^{- 4}

J

The answer is (C).

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option