Question

Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury T = 435.5 × 10^-3 Nm^-1

Answer 1

Radii of mercury droplets, r₁ = 0.1 cm = 1 × 10^-3 m

r₂ = 0.2 cm = 2 × 10^-3 m

Surface Tension, T = 435.5 × 10^-3 N/m

Let v₁ and v₂ be the volume of the droplets and V of the resulting drop, and R be the radius of big drop formed by collapsing.

Then V = v₁ + v₂ or \(\frac{4}{3}\)πR³ = \(\frac{4}{3}\)\(πr^3_1\) + \(\frac{4}{3}\)\(πr^3_2\)

Or R³ = \(r^3_1\) + \(r^3_2\) = (0.001 + 0.008) cm³ = 0.009 cm³

∴ R = 0.21 cm

Change in surface area, ∆A = 4π[R² − (\(r^2_1\) + \(r^2_2\))]

∴ Energy released, ∆U = T∆A = 4πT[R² − (\(r^2_1\) + \(r^2_2\))]

= 4 × 3.14 × 435.5 × 10^-3 [(0.21)² × 10^-4 – (1 × 10^-6 + 4 × 10^-6)]

= 435.5 × 4 × 3.14 [4.41 – 5] × 10^-6 × 10^-3

= −32.23 × 10^-7 J

∴ = 3.22 × 10^-6 J energy will be absorbed.