Question

A liquid of density \(800\,\text{kg m}^{-3}\) rises \(4\,\text{cm}\) in a capillary tube of inner radius \(0.2\,\text{mm}\). The capillary tube is dipped vertically in the liquid and the angle of contact between the capillary tube and the liquid is \(0^\circ\). The excess pressure inside the spherical drop of diameter \(2\,\text{cm}\) of the same liquid is \[ (\text{Acceleration due to gravity}=10\,\text{m s}^{-2}) \]

• A. \(9.6\,\text{N m}^{-2}\)
• B. \(12.8\,\text{N m}^{-2}\)
• C. \(6.4\,\text{N m}^{-2}\)
• D. \(3.2\,\text{N m}^{-2}\)

Hint:

Remember, \[ \boxed{ h=\frac{2T\cos\theta}{\rho gr} } \] for capillary rise, and \[ \boxed{ \Delta P=\frac{2T}{R} } \] for a liquid drop.

Answer 1

The Correct Option is C

Step 1: Determine the surface tension using capillary rise. For capillary rise, \[ h=\frac{2T\cos\theta}{\rho gr}. \] Since \[ \theta=0^\circ, \] \[ \cos\theta=1. \] Therefore, \[ T=\frac{h\rho gr}{2}. \] Substituting, \[ h=0.04\,\text{m}, \] \[ \rho=800\,\text{kg m}^{-3}, \] \[ g=10\,\text{m s}^{-2}, \] \[ r=0.2\,\text{mm}=2\times10^{-4}\,\text{m}, \] we obtain \[ T = \frac{0.04\times800\times10\times2\times10^{-4}}{2} = 0.032\,\text{N m}^{-1}. \]

Step 2: Use the formula for excess pressure in a liquid drop. For a spherical liquid drop, \[ \Delta P=\frac{2T}{R}. \] The diameter is \[ 2\,\text{cm}, \] so the radius is \[ R=1\,\text{cm}=0.01\,\text{m}. \] Hence, \[ \Delta P = \frac{2\times0.032}{0.01} = 6.4\,\text{N m}^{-2}. \] Therefore, \[ \boxed{\Delta P=6.4\,\text{N m}^{-2}.} \] Hence, the correct option is \(\boxed{(C)}\).