A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is . The value of is

Question

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is $α \times 10^{- 6} J$ . The value of $α$ is ______. (Take surface tension of liquid = 0.08 N/m) [2026]

Smart Achievers · Accepted Answer

(2)

$V_{1} = V_{2}$

$\frac{4}{3} π R^{3} = 512 (\frac{4}{3} π r^{3})$

$R = {(512)}^{1 / 3} r \Rightarrow r = \frac{1}{8} mm$

$Surface energy = S E_{2} - S E_{1}$

$Δ S E = 512 [T \times 4 π r^{2}] - [T \times 4 π R^{2}]$

$Δ S E = 4 π T [512 r^{2} - R^{2}]$

$Δ S E = 4 π (0.08) [512 {(\frac{1}{8})}^{2} - 1^{2}] \times 10^{- 6} \approx 7 \times 10^{- 6}$

$\Rightarrow α = 7$

Solution

Q.
A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is $α \times 10^{- 6} J$ . The value of $α$ is ______. (Take surface tension of liquid = 0.08 N/m) [2026]

1 10

2 7

3 8

4 11

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Solution

Q. A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α×10-6 J. The value of α is ______. (Take surface tension of liquid = 0.08 N/m) [2026]

1 10

2 7

3 8

4 11

Ans. (2) V1=V2 43πR3=512(43πr3) R=(512)1/3r⇒r=18 mm Surface energy=SE2-SE1 ΔSE=512[T×4πr2]-[T×4πR2] ΔSE=4πT[512r2-R2] ΔSE=4π(0.08)[512(18)2-12]×10-6≈7×10-6 ⇒α=7

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Q.
A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is $α \times 10^{- 6} J$ . The value of $α$ is ______. (Take surface tension of liquid = 0.08 N/m) [2026]