The frequency of sonometer wire is f. The frequency becomes `f//2` when the mass producing the tension is completely immersed in water and on immersing the mass in a certain liquid, frequency becomes `f//3`. The relative density of the liquid is

Options

A. 0.32
B. 1.67
C. 1.41
D. 1.18

Explanation

To solve the problem, we need to analyze the changes in frequency of the sonometer wire when the mass is immersed in water and then in a certain liquid. The relationship between frequency, tension, and linear density will guide us through the calculations.

### Step-by-Step Solution:

1. **Understanding the Frequency Formula:**
The frequency \( f \) of a sonometer wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire, \( L \) is the length of the wire, and \( \mu \) is the linear density of the wire.

2. **Initial Condition:**
Initially, the frequency is \( f \).

3. **Condition When Immersed in Water:**
When the mass is completely immersed in water, the frequency becomes \( \frac{f}{2} \). The tension \( T_w \) when immersed in water can be expressed as:
\[ T_w = mg - \text{upthrust} \]
The upthrust (buoyant force) is given by:
\[ \text{upthrust} = \rho_w V g \]
Therefore, the tension becomes:
\[ T_w = \rho V g - \rho_w V g = Vg(\rho - \rho_w) \]
Substituting this into the frequency formula gives:
\[ \frac{f}{2} = \frac{1}{2L} \sqrt{\frac{Vg(\rho - \rho_w)}{\mu}} \]

4. **Setting Up the Equation:**
Rearranging gives:
\[ \frac{f}{2} = \frac{1}{2L} \sqrt{\frac{Vg(\rho - \rho_w)}{\mu}} \implies 1 = \sqrt{\frac{Vg(\rho - \rho_w)}{\mu}} \cdot \frac{L}{f} \]
Squaring both sides:
\[ 1 = \frac{Vg(\rho - \rho_w)}{\mu} \cdot \left(\frac{L}{f}\right)^2 \]

5. **Condition When Immersed in a Liquid:**
When the mass is immersed in a certain liquid, the frequency becomes \( \frac{f}{3} \). The tension \( T_l \) in this case is:
\[ T_l = mg - \text{upthrust in liquid} = \rho V g - \rho_l V g = Vg(\rho - \rho_l) \]
Therefore, the frequency becomes:
\[ \frac{f}{3} = \frac{1}{2L} \sqrt{\frac{Vg(\rho - \rho_l)}{\mu}} \]

6. **Setting Up the Second Equation:**
Rearranging gives:
\[ \frac{f}{3} = \frac{1}{2L} \sqrt{\frac{Vg(\rho - \rho_l)}{\mu}} \implies 1 = \sqrt{\frac{Vg(\rho - \rho_l)}{\mu}} \cdot \frac{6L}{f} \]
Squaring both sides:
\[ 1 = \frac{Vg(\rho - \rho_l)}{\mu} \cdot \left(\frac{6L}{f}\right)^2 \]

7. **Taking Ratios:**
Now we take the ratio of the two equations derived from the frequencies in water and the liquid:
\[ \frac{\rho - \rho_l}{\rho - \rho_w} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]

8. **Substituting Known Values:**
We know from the first condition that:
\[ \rho - \rho_w = \frac{3}{4} \rho \]
Substituting this into the ratio gives:
\[ \frac{\rho - \rho_l}{\frac{3}{4} \rho} = \frac{4}{9} \]

9. **Solving for Relative Density:**
Rearranging gives:
\[ \rho - \rho_l = \frac{4}{9} \cdot \frac{3}{4} \rho = \frac{1}{3} \rho \]
Thus:
\[ \rho_l = \rho - \frac{1}{3} \rho = \frac{2}{3} \rho \]
Finally, the relative density of the liquid with respect to water is:
\[ \text{Relative Density} = \frac{\rho_l}{\rho_w} = \frac{\frac{2}{3} \rho}{\rho_w} \]
Since \( \rho = \frac{4}{3} \rho_w \):
\[ \text{Relative Density} = \frac{\frac{2}{3} \cdot \frac{4}{3} \rho_w}{\rho_w} = \frac{8}{9} \approx 1.18 \]

### Final Answer:
The relative density of the liquid is approximately **1.18**.