A resonance tube is resonated with tunning fork of frequency 256Hz. If the length of foirst and second resonating air coloumns are 32 cm and 100cm, then end correction will be

Options

A. 1cm
B. 2cm
C. 4cm
D. 6cm

Answer

To find the end correction (E) in a resonance tube experiment, we can follow these steps:

### Step 1: Understand the Resonance Conditions
In a resonance tube, the first resonating air column corresponds to a quarter wavelength (λ/4) and the second resonating air column corresponds to three-quarters of the wavelength (3λ/4).

### Step 2: Write the Equations for Resonating Columns
For the first resonating column (L1 = 32 cm):
\[ \frac{\lambda}{4} = L_1 + E \quad \text{(1)} \]
For the second resonating column (L2 = 100 cm):
\[ \frac{3\lambda}{4} = L_2 + E \quad \text{(2)} \]

### Step 3: Substitute the Lengths into the Equations
Substituting the lengths into the equations:
1. From Equation (1):
\[ \frac{\lambda}{4} = 32 + E \quad \text{(3)} \]
2. From Equation (2):
\[ \frac{3\lambda}{4} = 100 + E \quad \text{(4)} \]

### Step 4: Express λ from Equation (3)
From Equation (3), we can express λ:
\[ \lambda = 4(32 + E) = 128 + 4E \quad \text{(5)} \]

### Step 5: Substitute λ into Equation (4)
Now substitute λ from Equation (5) into Equation (4):
\[ \frac{3(128 + 4E)}{4} = 100 + E \]
This simplifies to:
\[ 3(128 + 4E) = 400 + 4E \]
\[ 384 + 12E = 400 + 4E \]

### Step 6: Solve for E
Rearranging the equation gives:
\[ 12E - 4E = 400 - 384 \]
\[ 8E = 16 \]
\[ E = 2 \text{ cm} \]

### Conclusion
The end correction (E) is **2 cm**.

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