Two capillary tubes of same diameter are put vertically one each in two liquids whose relative densities are 0.8 and 0.6 and surface tensions are 60 dyne/cm and 50 dyne/cm respectively. Ratio of heights of liquids in the two tubes `(h_(1))/(h_(2))` is

Options

A. `(10)/(9)`
B. `(3)/(10)`
C. `(10)/(3)`
D. `(9)/(10)`

Answer

To solve the problem of finding the ratio of heights of liquids in two capillary tubes, we can use the formula for capillary rise:

\[ h = \frac{2T \cos \theta}{\rho g r} \]

Where:
- \( h \) is the height of the liquid column,
- \( T \) is the surface tension of the liquid,
- \( \theta \) is the contact angle (which we can assume to be the same for both liquids),
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the capillary tube.

Given that the two capillary tubes have the same diameter, the radius \( r \) and the acceleration due to gravity \( g \) will cancel out when we take the ratio of the heights of the liquids in the two tubes.

### Step 1: Write the formula for the heights of the liquids in both tubes.

For liquid 1:
\[ h_1 = \frac{2T_1 \cos \theta}{\rho_1 g r} \]

For liquid 2:
\[ h_2 = \frac{2T_2 \cos \theta}{\rho_2 g r} \]

### Step 2: Take the ratio of the heights \( \frac{h_1}{h_2} \).

\[ \frac{h_1}{h_2} = \frac{\frac{2T_1 \cos \theta}{\rho_1 g r}}{\frac{2T_2 \cos \theta}{\rho_2 g r}} \]

### Step 3: Simplify the ratio.

Since \( 2 \), \( \cos \theta \), \( g \), and \( r \) are common in both the numerator and denominator, they cancel out:

\[ \frac{h_1}{h_2} = \frac{T_1}{T_2} \cdot \frac{\rho_2}{\rho_1} \]

### Step 4: Substitute the given values.

Given:
- \( T_1 = 60 \, \text{dyne/cm} \)
- \( T_2 = 50 \, \text{dyne/cm} \)
- \( \rho_1 = 0.8 \, \text{g/cm}^3 \)
- \( \rho_2 = 0.6 \, \text{g/cm}^3 \)

Substituting these values into the ratio:

\[ \frac{h_1}{h_2} = \frac{60}{50} \cdot \frac{0.6}{0.8} \]

### Step 5: Calculate the ratio.

Calculating the first part:

\[ \frac{60}{50} = \frac{6}{5} \]

Calculating the second part:

\[ \frac{0.6}{0.8} = \frac{6}{8} = \frac{3}{4} \]

Now multiply the two fractions:

\[ \frac{h_1}{h_2} = \frac{6}{5} \cdot \frac{3}{4} = \frac{18}{20} = \frac{9}{10} \]

### Final Result:

Thus, the ratio of heights of liquids in the two tubes is:

\[ \frac{h_1}{h_2} = \frac{9}{10} \]