Question

A capillary tube of diameter $1\times {10}^{-3}\text{m}$ is dipped in a container containing water. Given the surface tension of water is $0.072\frac{\text{N}}{\text{m}}$ and angle of contact is $0^{\circ }$, choose the correct option(s).

• A. The liquid level in the capillary will be more than that in the container.
• B. The liquid level in the capillary will be less than that in the container.
• C. The capillary rise will be $2.94\times {10}^{-2}\text{m}$.
• D. The capillary drop will be $1.14\times {10}^{-2}\text{m}$.

Answer 1

Answer: (A), (C)

The correct options are
A The liquid level in the capillary will be more than that in the container.
C The capillary rise will be $2.94\times {10}^{-2}\text{m}$.

We know the capillary rise is given by
$h=\frac{2Tcos\theta }{R\rho g}$
Since angle of contact is $0^o$, height will be positive and there will be capillary rise.
Putting the given values in the expression,
$h=\frac{2\times 0.072\times \cos 0^o}{5\times {10}^{-4}\times 1000\times 9.8}=2.938\times {10}^{-2}\text{m}$
Options (a) and (c) are correct.