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. `113.5cm`

Answer

To solve the problem of finding the end correction in a resonance pipe, we can follow these steps:

### Step 1: Understand the concept of resonance in a pipe
In a resonance pipe, the first and second resonance conditions occur at specific depths. The first resonance occurs at a depth corresponding to \( \frac{\lambda}{4} \) (where \( \lambda \) is the wavelength), and the second resonance occurs at \( \frac{3\lambda}{4} \).

### Step 2: Identify the given values
From the problem, we have:
- Depth for the first resonance (\( L_1 \)) = 22.7 cm
- Depth for the second resonance (\( L_2 \)) = 70.2 cm

### Step 3: Set up the relationship using the resonance formula
The relationship between the depths and the end correction (\( x \)) can be expressed as:
\[ \frac{L_1 + x}{L_2 + x} = \frac{1}{3} \]
This is derived from the fact that \( L_1 + x \) corresponds to \( \frac{\lambda}{4} \) and \( L_2 + x \) corresponds to \( \frac{3\lambda}{4} \).

### Step 4: Rearranging the formula
Cross-multiplying gives us:
\[ 3(L_1 + x) = L_2 + x \]
Expanding this, we have:
\[ 3L_1 + 3x = L_2 + x \]

### Step 5: Isolate \( x \)
Rearranging the equation to isolate \( x \):
\[ 3L_1 - L_2 = x - 3x \]
\[ 3L_1 - L_2 = -2x \]
\[ x = \frac{3L_1 - L_2}{2} \]

### Step 6: Substitute the known values
Substituting \( L_1 = 70.2 \) cm and \( L_2 = 22.7 \) cm into the equation:
\[ x = \frac{3(70.2) - 22.7}{2} \]
Calculating:
\[ x = \frac{210.6 - 22.7}{2} = \frac{187.9}{2} = 93.95 \text{ cm} \]

### Step 7: Calculate the end correction
The end correction \( x \) is:
\[ x = 93.95 \text{ cm} \]

### Final Answer
The end correction is approximately \( 1.05 \) cm.

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