Question

The pressure inside a mercury drop of diameter 2 mm is (Surface tension of mercury $= 0.5 Nm^{-1}$ and atmospheric pressure $= 10^5$ Pa)

• A. $1.020 \times 10^5$ Pa
• B. $1.005 \times 10^5$ Pa
• C. $1.100 \times 10^5$ Pa
• D. $1.010 \times 10^5$ Pa

Hint:

Drop vs Bubble: - Liquid Drop (in air): 1 Surface $\rightarrow \Delta P = 2T/R$ - Soap Bubble (in air): 2 Surfaces $\rightarrow \Delta P = 4T/R$ - Air Bubble (in liquid): 1 Surface $\rightarrow \Delta P = 2T/R$

Answer 1

The Correct Option is D

Step 1: Formula for Pressure inside a Drop:
A liquid drop has only one free surface. The excess pressure inside is given by: \[ P_{excess} = \frac{2T}{R} \] Total Pressure $P_{in} = P_{atm} + P_{excess}$.
Step 2: Given Values:
Diameter $D = 2$ mm $\Rightarrow$ Radius $R = 1$ mm $= 10^{-3}$ m. Surface Tension $T = 0.5 \, Nm^{-1}$. Atmospheric Pressure $P_{atm} = 10^5$ Pa.
Step 3: Calculation:
\[ P_{excess} = \frac{2 \times 0.5}{10^{-3}} = \frac{1}{10^{-3}} = 1000 \, \text{Pa} = 10^3 \, \text{Pa} \] Convert excess pressure to match the order of atmospheric pressure ($10^5$): $P_{excess} = 0.01 \times 10^5$ Pa. \[ P_{in} = 10^5 + 0.01 \times 10^5 = (1 + 0.01) \times 10^5 \] \[ P_{in} = 1.01 \times 10^5 \, \text{Pa} = 1.010 \times 10^5 \, \text{Pa} \]