Q.25 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\(^2\))
Best answer
Answer: Pressure inside \(= 1.0178 \times 10^5\) N/m\(^2\) (excess pressure 480 N/m\(^2\)).
Given: \(r = 0.3\) mm \(= 3 \times 10^{-4}\) m, \(T = 72\) dyne/cm \(= 72 \times 10^{-3}\) N/m, \(P_o = 1.013 \times 10^5\) N/m\(^2\).
A liquid drop has one free surface, so the excess pressure inside it is
\[P_i - P_o = \frac{2T}{r} = \frac{2 \times 72 \times 10^{-3}}{3 \times 10^{-4}} = 480 \text{ N/m}^2\]Pressure inside the drop:
\[P_i = 1.013 \times 10^5 + 480 = 101300 + 480 = 101780 \text{ N/m}^2 = 1.0178 \times 10^5 \text{ N/m}^2\]Log in or create a free account to answer this question.
