A uniform string resonates with a tuning fork, at a maximum tension of 32 N. If it is divided into two segments by placing a wedge at a distance one fourth of length from one end, then resonance frequency will occur at a maximum value of tension :-

Options

A. 2N
B. 4N
C. 8N
D. 16N

Answer

To solve the problem, we need to determine the maximum tension under which the divided segments of the string will resonate. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have a uniform string of length \( L \) that resonates with a tuning fork at a maximum tension \( T_1 = 32 \, \text{N} \). The string is divided into two segments by placing a wedge at a distance of \( \frac{1}{4}L \) from one end.

### Step 2: Identify the Lengths
Let:
- \( L_1 \) = Total length of the string = \( L \)
- \( L_2 \) = Length of the segment on one side of the wedge = \( \frac{1}{4}L \)

### Step 3: Use the Formula for Frequency
The fundamental frequency \( f \) of a string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension and \( \mu \) is the linear mass density of the string.

### Step 4: Relate the Lengths and Tensions
From the frequency formula, we can derive that:
\[ L \propto \sqrt{T} \]
This means:
\[ \frac{L_1}{L_2} = \frac{L}{\frac{1}{4}L} = 4 \]
Thus, we can express this relationship in terms of tensions:
\[ \frac{L_1}{L_2} = \sqrt{\frac{T_1}{T_2}} \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:
\[ 4 = \sqrt{\frac{32}{T_2}} \]

### Step 6: Square Both Sides
Squaring both sides gives us:
\[ 16 = \frac{32}{T_2} \]

### Step 7: Solve for \( T_2 \)
Rearranging the equation to solve for \( T_2 \):
\[ T_2 = \frac{32}{16} = 2 \, \text{N} \]

### Conclusion
The maximum tension under which the segment will resonate is \( T_2 = 2 \, \text{N} \).

---