A liquid drop of diameter 2 mm breaks into 512 identical smaller droplets. The surface tension of the liquid is 0.08 N·m⁻¹. If the increase in surface energy is α×10⁻⁶ J, what is α to the nearest integer?
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Question

A liquid drop of diameter 2 mm breaks into 512 identical smaller droplets. The surface tension of the liquid is 0.08 N·m⁻¹. If the increase in surface energy is α×10⁻⁶ J, what is α to the nearest integer?

Options

G1 Physics · Accepted Answer

The original drop radius is R = 1 mm = 1×10⁻³ m. Let the radius of each small droplet be r. Conservation of volume gives (4 ÷ 3)πR³ = 512(4 ÷ 3)πr³. Therefore, R³ = 512r³. Since 512 = 8³, R = 8r, so r = R ÷ 8. Initial surface area: A₁ = 4πR². Final total surface area: A₂ = 512×4πr² = 512×4π(R ÷ 8)² = 512×4πR² ÷ 64 = 8×4πR². Therefore, ΔA = A₂-A₁ = 7×4πR². Increase in surface energy: ΔU = TΔA = 0.08×7×4π×(10⁻³)² ≈ 7.04×10⁻⁶ J. Thus, α ≈ 7. Correct option: B.