What is the pressure inside a drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 0.465N/m. The atmospheric pressure is 1.01 x10⁵ Pa. Also give the excess pressure inside the drop.

Solution

Given,

Radius of drop of mercury, r = 3.0 mm

Surface tension of mercury = 0.465 N/m

Atmospheric pressure, P_A = 1.01

\times

10⁵ Pa

Excess of pressure inside the liquid drop is given by,

p = \frac{2 T}{R} Here, T = 0.465 N / m R = 3 mm = 3 \times 10^{- 3} m ∴ E x c e s s p r e s s u r e, p = \frac{2 \times 4.65 \times 10^{- 1}}{3 \times 10^{- 3}} = 310 Pa = 0.0031 \times 10^{5} Pa ∴ Pressure inside the drop = P_{0} + p = 1.01 x 10^{5} + 0.0031 \times 10^{5} Pa = 1.0131 \times 10^{5} Pa

Some More Questions From Mechanical Properties of Fluids Chapter

For Daily Free Study Material Join wiredfaculty