Question

A mercury drop of radius 1 cm is sprayed into 10⁶ drops of equal size. The energy expressed in joules is (surface tension of mercury is 460 × 10⁻3N/m):

• A. 0.057
• B. 5.7
• C. 5.7 × 10⁻4
• D. 5.7 × 10⁻6

Hint:

When a liquid drop breaks into smaller drops, surface area increases and energy required is: E = T (Af - Aᵢ) Radius of each new drop is found using volume conservation.

Answer 1

The Correct Option is A

Step 1: Radius of original drop R = 1 cm = 10⁻2m Initial surface area: Aᵢ = 4π R² = 4π (10⁻2)² = 4π × 10⁻4 

Step 2: Number of drops n = 10⁶ Radius of each small drop: r = fracRn¹/3 = frac10⁻2100 = 10⁻4m Final surface area: Af = n · 4π r² = 10⁶ · 4π (10⁻4)² = 4π × 10⁻2 

Step 3: Increase in surface area Δ A = Af - Aᵢ = 4π (10⁻2 - 10⁻4) 

Step 4: Increase in surface energy E = T Δ A = 0.46 × 4π × 0.0099 ≈ 0.057 J