Topic: Fluid Mechanics
A spherical water drop of diameter 2 mm splits in to 8 small drops of equal size in air at temperature of 25 °C. The surface tension coefficient of water in air is 0.073 N/m and dynamic viscosity is 0.01 poise. The approximate work done in splitting up the drop is
- A. 0.29 x 10^-6 J
- B. 0.92 x 10^-6 J
- C. 0.86 x 10^-6 J
- D. 0.78 x 10^-6 J
Answer: B. 0.92 x 10^-6 J
Explanation: Initial volume = Final volume, so (4/3)*pi*R³ = 8 * (4/3)*pi*r³, which means r = R/2. The initial radius R = 1 mm. Change in surface area = Final Area - Initial Area = 8*(4*pi*r²) - 4*pi*R² = 8*4*pi*(R/2)^2 - 4*pi*R² = 4*pi*R². Work done = Surface tension * Change in area = 0.073 * 4 * pi * (10^-3)^2 = 0.917 x 10^-6 J, which is approximately 0.92 x 10^-6 J.