Question

Water rises to a height of \(15\,\text{mm}\) in a capillary tube having cross-sectional area \(A\). If the cross-sectional area of the tube is made \( \dfrac{A}{3} \), then the water will rise to a height

• A. \(15\sqrt{3}\times10^{-3}\,\text{m}\)
• B. \(20\sqrt{3}\times10^{-3}\,\text{m}\)
• C. \(5\sqrt{3}\times10^{-3}\,\text{m}\)
• D. \(10\sqrt{3}\times10^{-3}\,\text{m}\)

Hint:

Capillary rise varies inversely with the square root of cross-sectional area of the tube.

Answer 1

The Correct Option is A

Step 1: Relation for capillary rise.
Capillary rise is given by \[ h = \frac{2T\cos\theta}{\rho g r} \] Thus, \[ h \propto \frac{1}{r} \]

Step 2: Relate radius with cross-sectional area.
\[ A = \pi r^2 \Rightarrow r \propto \sqrt{A} \]

Step 3: Find relation between height and area.
\[ h \propto \frac{1}{\sqrt{A}} \]

Step 4: Apply given change in area.
If \(A' = \dfrac{A}{3}\), then \[ h' = h\sqrt{3} \] \[ h' = 15\,\text{mm}\times\sqrt{3} = 15\sqrt{3}\times10^{-3}\,\text{m} \]