In a resonance pipe the first and second resonance are obtained at depths 22.7 cm and 70.2 respectively. What will be the end correction?

Options

A. 1.05 cm
B. 115.5 cm
C. 92.5 cm
D. 113.5 cm

Answer

To find the end correction in a resonance pipe, we can follow these steps:

### Step 1: Understand the Resonance Condition
In a resonance pipe, the lengths at which resonance occurs correspond to odd multiples of a quarter wavelength. The first resonance occurs at \( L_1 \) and the second resonance occurs at \( L_2 \).

### Step 2: Identify the Given Values
From the problem:
- First resonance depth \( L_1 = 22.7 \, \text{cm} \)
- Second resonance depth \( L_2 = 70.2 \, \text{cm} \)

### Step 3: Apply the Resonance Condition
The lengths at which the resonances occur can be expressed as:
- For the first resonance: \( L_1 + x = \frac{\lambda}{4} \)
- For the second resonance: \( L_2 + x = \frac{3\lambda}{4} \)

Where \( x \) is the end correction.

### Step 4: Set Up the Equations
From the above expressions, we can derive:
1. \( L_1 + x = \frac{\lambda}{4} \) (1)
2. \( L_2 + x = \frac{3\lambda}{4} \) (2)

### Step 5: Eliminate \( \lambda \)
Subtract equation (1) from equation (2):
\[ (L_2 + x) - (L_1 + x) = \frac{3\lambda}{4} - \frac{\lambda}{4} \]
This simplifies to:
\[ L_2 - L_1 = \frac{2\lambda}{4} = \frac{\lambda}{2} \]
From this, we can express \( \lambda \):
\[ \lambda = 2(L_2 - L_1) \]

### Step 6: Substitute Back to Find \( x \)
Now substitute \( \lambda \) back into either equation (1) or (2). We'll use equation (1):
\[ L_1 + x = \frac{2(L_2 - L_1)}{4} \]
This simplifies to:
\[ L_1 + x = \frac{L_2 - L_1}{2} \]
Rearranging gives:
\[ x = \frac{L_2 - L_1}{2} - L_1 \]
\[ x = \frac{L_2 - 3L_1}{2} \]

### Step 7: Plug in the Values
Substituting the values of \( L_1 \) and \( L_2 \):
\[ x = \frac{70.2 - 3 \times 22.7}{2} \]
Calculating:
\[ x = \frac{70.2 - 68.1}{2} = \frac{2.1}{2} = 1.05 \, \text{cm} \]

### Final Answer
The end correction \( x \) is \( 1.05 \, \text{cm} \).

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