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`

Explanation

To find the end correction in a resonance pipe given the first and second resonance depths, we can follow these steps:

### Step 1: Understand the Resonance Conditions
In a closed pipe, the first resonance occurs at a depth where the length of the air column corresponds to a quarter of the wavelength (λ/4), and the second resonance occurs at a depth corresponding to three-quarters of the wavelength (3λ/4).

### Step 2: Define the Variables
Let:
- \( L_1 = 22.7 \, \text{cm} \) (depth for first resonance)
- \( L_2 = 70.2 \, \text{cm} \) (depth for second resonance)
- \( X \) = end correction

### Step 3: Relate the Lengths and Wavelengths
From the resonance conditions:
- For the first resonance: \( L_1 + X = \frac{\lambda}{4} \)
- For the second resonance: \( L_2 + X = \frac{3\lambda}{4} \)

### Step 4: Set Up the Equations
We can express the two equations as:
1. \( L_1 + X = \frac{\lambda}{4} \) (1)
2. \( L_2 + X = \frac{3\lambda}{4} \) (2)

### Step 5: Eliminate λ
From equation (1), we can express λ:
\[ \lambda = 4(L_1 + X) \]
Substituting this into equation (2):
\[ L_2 + X = \frac{3}{4} \times 4(L_1 + X) \]
This simplifies to:
\[ L_2 + X = 3(L_1 + X) \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ L_2 + X = 3L_1 + 3X \]
\[ L_2 - 3L_1 = 2X \]

### Step 7: Solve for X
Now we can substitute the values of \( L_1 \) and \( L_2 \):
\[ X = \frac{L_2 - 3L_1}{2} \]
\[ X = \frac{70.2 - 3 \times 22.7}{2} \]
Calculating this:
\[ X = \frac{70.2 - 68.1}{2} = \frac{2.1}{2} = 1.05 \, \text{cm} \]

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

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