Question

Diameter of a water drop is 0.6 mm. Calculate the pressure inside a liquid drop.

(T = 72 dyne/cm, atmospheric pressure = 1.013 × 10⁵ N/m²)

Answer 1

Given:

Diameter of a water drop = 0.6 mm = 0.06 cm

Radius (r) = `0.06/2 = 0.03`

Surface tension of water (T) = 72 dyne/cm = 72 × 10⁻³ N/m

Atomospheric pressure = 1.013 × 10⁵ N/m²

Formula:

For a liquid drop, the excess pressure inside is given by:

ΔP = `(2T)/r`

where:

• ΔP = excess pressure inside the drop
• T = surface tension
• r = radius of the drop

Calculation:

ΔP = `(2 xx (72 xx 10^(-3)))/(0.03 xx 10^(-2))`

ΔP = `(144 xx 10^(-3))/(0.03 xx 10^(-2))`

= `(144 xx 10^(-3) xx 10^2)/(0.03)`

= `(144 xx 10^(-1))/(3 xx 10^(-2))`

= `(144 xx 10^(-1) xx 10^2)/3`

= `(144 xx 10)/3`

= `1440/3`

ΔP = 480 N/m²

The total pressure inside the drop is:

P = P_atm + ΔP

P = (1.013 × 10⁵) + 480

= 101300 + 480

= 101780

= 1.0178 × 10⁵ N/m²

The total pressure inside the liquid drop is 1.0178 × 10⁵ N/m².