Question

Diameter of a water drop is 0.6 mm. Calculate the pressure inside a liquid drop. (T = 72 dyne/cm, atmospheric pressure = \(1.013 \times 10^5\) N/m\(^2\))

Hint:

Remember the formula has a factor of 2 for a drop (\(2T/r\)) because there is one surface, but a factor of 4 for a soap bubble (\(4T/r\)) because it has two surfaces (inner and outer).

Answer 1

First, we calculate the excess pressure (\(P_{excess}\)) inside the water drop due to surface tension. The formula for a liquid drop is: \[ P_{excess} = \frac{2T}{r} \] We need to convert the given values to SI units.
Diameter = 0.6 mm, so radius \(r = 0.3 \, mm = 0.3 \times 10^{-3} \, m\)
Surface tension, \(T = 72 \, \text{dyne/cm}\).
Since \(1 \, \text{dyne} = 10^{-5} \, N\) and \(1 \, \text{cm} = 10^{-2} \, m\),
\(T = \frac{72 \times 10^{-5} \, N}{10^{-2} \, m} = 72 \times 10^{-3} \, N/m\)
Now, calculate the excess pressure: \[ P_{excess} = \frac{2 \times (72 \times 10^{-3})}{0.3 \times 10^{-3}} = \frac{144}{0.3} = 480 \, N/m^2 \text{ (or Pa)} \] The total pressure inside the drop (\(P_{inside}\)) is the sum of the atmospheric pressure (\(P_{atm}\)) and the excess pressure. \[ P_{inside} = P_{atm} + P_{excess} \] \[ P_{inside} = (1.013 \times 10^5) + 480 \] \[ P_{inside} = 101300 + 480 = 101780 \, N/m^2 \] The pressure inside the water drop is \(1.0178 \times 10^5\) N/m\(^2\).