A string under a tension of `100 N` , emitting its fundamental mode , gives `5 beats//s` with a tuning fork. When the tension is increased to `121 N`, again `5 beats//s` are heard. The frequency of the fork is

Options

A. `105 Hz`
B. `95 Hz`
C. `210 Hz`
D. `190 Hz`

Explanation

To solve the problem, we need to analyze the relationship between the tension in the string, the frequency of the string, and the frequency of the tuning fork. Here are the steps to derive the solution:

### Step 1: Understand the relationship between tension and frequency
The frequency of a vibrating string is given by the formula:
\[ f \propto \sqrt{T} \]
where \( f \) is the frequency and \( T \) is the tension in the string.

### Step 2: Set up the equations for both tensions
Let \( f_1 \) be the frequency of the string under a tension of \( 100 \, \text{N} \) and \( f_2 \) be the frequency of the string under a tension of \( 121 \, \text{N} \). We can express these frequencies as:
\[ f_1 = k \sqrt{100} = 10k \]
\[ f_2 = k \sqrt{121} = 11k \]
where \( k \) is a constant of proportionality.

### Step 3: Set up the equations for beats
We know that beats occur when there is a difference in frequencies. The frequency of the tuning fork is \( f \). Given that there are \( 5 \) beats per second in both cases, we can set up the following equations:
1. For the first case (tension \( 100 \, \text{N} \)):
\[ |f - f_1| = 5 \]
This gives us two cases:
\[ f - f_1 = 5 \quad \text{(1)} \]
or
\[ f_1 - f = 5 \quad \text{(2)} \]

2. For the second case (tension \( 121 \, \text{N} \)):
\[ |f - f_2| = 5 \]
This also gives us two cases:
\[ f - f_2 = 5 \quad \text{(3)} \]
or
\[ f_2 - f = 5 \quad \text{(4)} \]

### Step 4: Solve the equations
Assuming case (1) and (4):
From (1):
\[ f = f_1 + 5 = 10k + 5 \]
From (4):
\[ f_2 - f = 5 \implies 11k - f = 5 \implies f = 11k - 5 \]

Setting the two expressions for \( f \) equal to each other:
\[ 10k + 5 = 11k - 5 \]
Solving for \( k \):
\[ 5 + 5 = 11k - 10k \implies 10 = k \]

### Step 5: Find the frequency of the tuning fork
Now substituting \( k \) back into the equation for \( f \):
\[ f = 10k + 5 = 10(10) + 5 = 100 + 5 = 105 \, \text{Hz} \]

### Conclusion
The frequency of the tuning fork is:
\[ \boxed{105 \, \text{Hz}} \]