An open pipe of length ` 2m` is dipped in water . To what depth ` x` is to be immersed in water so that it may resonate with a tuning fork of frequency `170 Hz` when vibrating in its overtone . Speed of sound in air is `340 m//s`

Options

A. `0.5 m`
B. `0.75 m`
C. `1 m`
D. `1.5 m`

Explanation

To solve the problem step by step, we will follow the physics principles related to sound waves in open pipes and their resonant frequencies.

### Step 1: Understand the Problem
We have an open pipe of length \( L = 2 \, \text{m} \) that is partially submerged in water. We need to find the depth \( x \) to which the pipe is immersed so that it resonates with a tuning fork of frequency \( f = 170 \, \text{Hz} \) in its first overtone.

### Step 2: Identify the Type of Resonance
In an open pipe, the fundamental frequency and overtones can be expressed in terms of the length of the pipe that is open to air. Since the pipe is partially submerged, we need to find the effective length of the pipe that is open to air, denoted as \( L_1 \).

### Step 3: Frequency of the First Overtone
The frequency of the first overtone (which is the second harmonic) for an open pipe is given by the formula:
\[ f = \frac{3V}{4L_1} \]
where \( V \) is the speed of sound in air, and \( L_1 \) is the length of the pipe that is above the water.

### Step 4: Substitute Known Values
We know:
- Speed of sound \( V = 340 \, \text{m/s} \)
- Frequency \( f = 170 \, \text{Hz} \)

Now, we can rearrange the formula to solve for \( L_1 \):
\[ L_1 = \frac{3V}{4f} \]

Substituting the known values:
\[ L_1 = \frac{3 \times 340}{4 \times 170} \]

### Step 5: Calculate \( L_1 \)
Calculating \( L_1 \):
\[ L_1 = \frac{1020}{680} = 1.5 \, \text{m} \]

### Step 6: Find the Depth \( x \)
Since the total length of the pipe is \( L = 2 \, \text{m} \) and the length above water is \( L_1 = 1.5 \, \text{m} \), the depth \( x \) to which the pipe is immersed in water is:
\[ x = L - L_1 = 2 - 1.5 = 0.5 \, \text{m} \]

### Final Answer
The depth \( x \) to which the pipe is immersed in water is \( 0.5 \, \text{m} \).

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