JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Fluid Mechanics

If 1000 droplets of water of surface tension

0.07

N/m. having same radius 1 mm each, combine to from a single drop. In the process the released surface energy is-
(Take

π = \frac{22}{7}

)

Answer: (B)

Solution

Volume is conserved when the drops merge,

1000 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}

, giving

R = (1000)^{1/3} r = 10 r = 10

mm. The surface energy released equals the surface tension times the decrease in surface area,

Δ E = T 4 π (1000 r^{2} - R^{2})

. With

r = 1 0^{- 3}

m and

R = 1 0^{- 2}

m this is

Δ E = 0.07 \times 4 \times \frac{22}{7} \times (1 0^{- 3} - 1 0^{- 4}) = 0.07 \times 4 \times \frac{22}{7} \times 9 \times 1 0^{- 4} = 7.92 \times 1 0^{- 4}

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option