A sonometer consists of two wire of length , same material whose radii are in the ratio `2:3`. The ratio of tension in two wire if their fundamental freqencies are equal is

Options

A. `1:4`
B. `2:3`
C. `9:4`
D. `4:9`

Explanation

To solve the problem, we need to find the ratio of the tensions in two wires of a sonometer, given that their fundamental frequencies are equal and their radii are in the ratio of 2:3.

### Step-by-Step Solution:

1. **Understanding the Fundamental Frequency Formula**:
The fundamental frequency (ν) of a wire is given by the formula:
\[ \nu = \frac{1}{2L} \sqrt{\frac{T}{M}} \]
where:
- \( \nu \) = fundamental frequency
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( M \) = mass of the wire

2. **Expressing Mass in Terms of Density and Volume**:
The mass \( M \) of the wire can be expressed as:
\[ M = \rho \cdot V \]
where \( \rho \) is the density and \( V \) is the volume of the wire. The volume \( V \) of a cylindrical wire is given by:
\[ V = \pi r^2 L \]
Therefore, we can write:
\[ M = \rho \cdot \pi r^2 L \]

3. **Substituting Mass into the Frequency Formula**:
Substituting \( M \) into the frequency formula gives:
\[ \nu = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot \pi r^2 L}} \]
Simplifying this, we get:
\[ \nu = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot \pi}} \cdot \frac{1}{\sqrt{L}} \cdot \frac{1}{r} \]
This shows that the frequency is directly proportional to \( \sqrt{T} \) and inversely proportional to \( r \).

4. **Setting Up the Ratio of Frequencies**:
Since the fundamental frequencies of both wires are equal, we can write:
\[ \frac{\nu_1}{\nu_2} = 1 \]
This implies:
\[ \frac{\sqrt{T_1}}{r_1} = \frac{\sqrt{T_2}}{r_2} \]

5. **Cross Multiplying**:
Cross multiplying gives:
\[ r_1 \sqrt{T_2} = r_2 \sqrt{T_1} \]

6. **Squaring Both Sides**:
Squaring both sides results in:
\[ r_1^2 T_2 = r_2^2 T_1 \]

7. **Finding the Ratio of Tensions**:
Rearranging gives:
\[ \frac{T_1}{T_2} = \frac{r_1^2}{r_2^2} \]

8. **Substituting the Ratio of Radii**:
Given that the ratio of the radii \( r_1:r_2 = 2:3 \), we can substitute:
\[ \frac{T_1}{T_2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]

9. **Final Answer**:
Thus, the ratio of the tensions in the two wires is:
\[ T_1 : T_2 = 4 : 9 \]