If the surface tension of water-air interface is 0.073 N/m, the gauge pressure inside a rain drop of 1mm diameter will be
- A. 0.146 N/m$^2$
- B. 73 N/m$^2$
- C. 146 N/m$^2$
- D. 292 N/m$^2$
Answer: D. 292 N/m$^2$
Explanation: Pressure inside a droplet $\Delta P = 4\sigma/d = 4(0.073)/0.001 = 292$ Pa.