In a resonance tube experiment ,a tuning fork resonates with an air column produces first resonating length of 12.5cm and second resonating length of 40cm . Then the end correction is

Options

A. 1cm
B. 1.25cm
C. 3cm
D. 0.12cm

Explanation

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

### Step 1: Understand the Concept of Resonating Lengths
In a resonance tube experiment, the first resonating length (L1) corresponds to the first harmonic (or fundamental frequency), while the second resonating length (L2) corresponds to the third harmonic. The end correction (E) accounts for the fact that the antinode of the wave does not occur exactly at the open end of the tube.

### Step 2: Write Down the Given Values
From the problem, we have:
- First resonating length (L1) = 12.5 cm
- Second resonating length (L2) = 40 cm

### Step 3: Set Up the Equations
For the first resonating length, we can express it as:
\[ L_1 = L + E \]
Where L is the effective length of the air column.

For the second resonating length, which corresponds to the third harmonic, we can express it as:
\[ L_2 = 3L + E \]

### Step 4: Substitute the Known Values
Now substituting the values we have:
1. From the first resonating length:
\[ 12.5 = L + E \] (Equation 1)

2. From the second resonating length:
\[ 40 = 3L + E \] (Equation 2)

### Step 5: Solve the Equations
From Equation 1, we can express L in terms of E:
\[ L = 12.5 - E \]

Now substitute this expression for L into Equation 2:
\[ 40 = 3(12.5 - E) + E \]

Expanding this gives:
\[ 40 = 37.5 - 3E + E \]
\[ 40 = 37.5 - 2E \]

Now, rearranging the equation to solve for E:
\[ 2E = 37.5 - 40 \]
\[ 2E = -2.5 \]
\[ E = -1.25 \text{ cm} \]

### Step 6: Correct the Sign
Since the end correction cannot be negative, we take the absolute value:
\[ E = 1.25 \text{ cm} \]

### Final Answer
The end correction is:
\[ \boxed{1.25 \text{ cm}} \]

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