A sonometer wire vibrating in its fundamental mode, is in unison with a tuning fork when the length of the wire between the bridges is 25 cm. By keeping the same tension , the length between the bridges is increased to 75 cm. The tuning fork can still be in resonance with the vibrating wire , provided the wire vibrates with

Options

A. 2 loops
B. 4 loops
C. 5 loops
D. 3 loops

Explanation

To solve the problem step-by-step, we will analyze the relationship between the frequency of the vibrating wire and its length in the fundamental mode.

### Step 1: Understand the relationship between frequency, tension, and length
In the fundamental mode of vibration, the frequency \( f \) of a vibrating string (or wire) is given by the formula:
\[ f = \frac{v}{2L} \]
where:
- \( v \) is the speed of the wave in the wire,
- \( L \) is the length of the wire between the bridges.

### Step 2: Set up the initial conditions
Initially, the length of the wire \( L_1 \) is 25 cm. The frequency at this length can be expressed as:
\[ f_1 = \frac{v}{2L_1} = \frac{v}{2 \times 25} \]

### Step 3: Change the length of the wire
Now, the length of the wire is increased to \( L_2 = 75 \) cm. The frequency at this new length can be expressed as:
\[ f_2 = \frac{v}{2L_2} = \frac{v}{2 \times 75} \]

### Step 4: Set the frequencies equal for resonance
Since the tuning fork can still be in resonance with the vibrating wire, we set \( f_1 = f_2 \):
\[ \frac{v}{2 \times 25} = \frac{v}{2 \times 75} \]

### Step 5: Simplify the equation
By canceling \( v \) and \( 2 \) from both sides, we get:
\[ \frac{1}{25} = \frac{1}{75} \]

### Step 6: Solve for the number of loops
To find the number of loops \( n \) that the wire can vibrate in at the new length, we use the relationship between the lengths:
\[ n = \frac{L_2}{L_1} = \frac{75}{25} = 3 \]

### Final Answer
Thus, the wire can vibrate with **3 loops** when the length is increased to 75 cm.

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