If tension in a sonometer wire is increased by 50% then fundamental frequency of the wire will be increased by 10 Hz. Assuming length as invariable, the frequency of the wire will be

Options

A. 45.4 Hz
B. 55.4 Hz
C. 65.4 Hz
D. 70 Hz

Explanation

To solve the problem step-by-step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the relationship between tension and frequency
The fundamental frequency \( \nu \) of a sonometer wire is given by the formula:
\[ \nu = \frac{1}{2L} \sqrt{\frac{T}{m}} \]
where:
- \( L \) is the length of the wire (assumed constant),
- \( T \) is the tension in the wire,
- \( m \) is the mass per unit length of the wire.

### Step 2: Define the initial frequency
Let the initial tension be \( T \) and the initial frequency be \( \nu_0 \):
\[ \nu_0 = \frac{1}{2L} \sqrt{\frac{T}{m}} \]

### Step 3: Calculate the new tension
If the tension is increased by 50%, the new tension \( T' \) can be expressed as:
\[ T' = T + 0.5T = 1.5T \]

### Step 4: Define the new frequency
The new frequency \( \nu' \) with the increased tension is given by:
\[ \nu' = \frac{1}{2L} \sqrt{\frac{T'}{m}} = \frac{1}{2L} \sqrt{\frac{1.5T}{m}} \]

### Step 5: Relate the new frequency to the old frequency
We can express \( \nu' \) in terms of \( \nu_0 \):
\[ \nu' = \frac{1}{2L} \sqrt{\frac{1.5T}{m}} = \sqrt{1.5} \cdot \frac{1}{2L} \sqrt{\frac{T}{m}} = \sqrt{1.5} \cdot \nu_0 \]
Calculating \( \sqrt{1.5} \):
\[ \sqrt{1.5} \approx 1.2247 \]
Thus,
\[ \nu' \approx 1.2247 \nu_0 \]

### Step 6: Set up the equation based on the problem statement
According to the problem, the increase in frequency is 10 Hz:
\[ \nu' - \nu_0 = 10 \text{ Hz} \]
Substituting the expression for \( \nu' \):
\[ 1.2247 \nu_0 - \nu_0 = 10 \]
This simplifies to:
\[ 0.2247 \nu_0 = 10 \]

### Step 7: Solve for \( \nu_0 \)
Now, we can solve for \( \nu_0 \):
\[ \nu_0 = \frac{10}{0.2247} \approx 44.5 \text{ Hz} \]

### Final Answer
The fundamental frequency of the wire is approximately:
\[ \nu_0 \approx 44.5 \text{ Hz} \]