Question

In a capillary tube having area of cross-section \(A\), water rises to a height \(h\). If cross-sectional area is reduced to \(\frac{A}{9}\), the rise of water in the capillary tube is

• A. \(3h\)
• B. \(9h\)
• C. \(h\)
• D. \(6h\)

Hint:

Capillary rise \(h \propto 1/r\). Area \(\propto r^2\), so \(h \propto 1/\sqrt{A}\). When area becomes \(A/9\), radius becomes \(r/3\), height becomes \(3h\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Capillary rise height \(h\) is inversely proportional to the radius of the tube. Area \(A = \pi r^2\), so \(r \propto \sqrt{A}\).

Step 2: Key Formula or Approach:
Height \(h = \frac{2S \cos \theta}{\rho g r}\), so \(h \propto \frac{1}{r}\).

Step 3: Detailed Explanation:
Initial radius \(r_1\), area \(A_1 = \pi r_1^2\). New area \(A_2 = \frac{A_1}{9} = \pi r_2^2\). Therefore \(r_2 = \frac{r_1}{3}\).
Since \(h \propto \frac{1}{r}\), \(h_2 = h_1 \times \frac{r_1}{r_2} = h \times 3 = 3h\).

Step 4: Final Answer:
Option (A) is correct.