Question

Two capillary tubes of same diameter are kept vertically in two liquids whose densities are in the ratio \( 4:3 \). If their surface tensions are in the ratio \( 6:5 \), the ratio of heights \( \left( \frac{h_1}{h_2} \right) \) of liquids in the two capillary tubes is (Their angle of contacts are same)

• A. \( 10 : 7 \)
• B. \( 8 : 7 \)
• C. \( 2 : 1 \)
• D. \( 3 : 2 \)

Hint:

The height of the liquid in the capillary tube is inversely proportional to the density of the liquid and directly proportional to the surface tension.

Answer 1

The Correct Option is A

Step 1: Formula for capillary rise.
The height of liquid in a capillary tube is given by the formula: \[ h = \frac{2T \cos \theta}{r \rho g}, \] where: - \( T \) is the surface tension,
- \( \theta \) is the angle of contact (same for both tubes),
- \( r \) is the radius of the capillary,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity.

Step 2: Ratio of heights of liquids.
The ratio of the heights of liquids in two capillary tubes is:
\[ \frac{h_1}{h_2} = \frac{\frac{2T_1 \cos \theta}{r \rho_1 g}}{\frac{2T_2 \cos \theta}{r \rho_2 g}} = \frac{T_1 \rho_2}{T_2 \rho_1}. \]

Step 3: Substituting the given values.
We are given that the surface tensions are in the ratio \( \frac{T_1}{T_2} = \frac{6}{5} \) and the densities are in the ratio \( \frac{\rho_1}{\rho_2} = \frac{4}{3} \). Substituting these values:
\[ \frac{h_1}{h_2} = \frac{6}{5} \times \frac{3}{4} = \frac{18}{20} = \frac{9}{10}. \]
Final Answer:
Thus, the ratio of the heights is:
\[ \boxed{10 : 7}. \]