Question

In a resonance tube open at one end, the end correction is \(1.1\) cm. If the shortest length of resonating air column with a tuning fork is \(18\) cm, the next resonating length will be

• A. \(45.9\) cm
• B. \(49.6\) cm
• C. \(51.3\) cm
• D. \(56.2\) cm

Hint:

In a tube closed at one end, resonances occur at lengths with end correction equal to lambda/4, 3 lambda/4 and so on.

Answer 1

The Correct Option is D

Step 1: Use the gap between resonances
Successive resonating lengths of a closed pipe differ by $\lambda/2$.

Step 2: Find lambda
$\dfrac{\lambda}{4} = L_1 + e = 19.1$ cm, so $\lambda = 76.4$ cm and $\dfrac{\lambda}{2} = 38.2$ cm.

Step 3: Add
$L_2 = L_1 + \dfrac{\lambda}{2} = 18 + 38.2 = 56.2$ cm.

Step 4: Check
$L_2 + e = 57.3 = 3\times19.1$, which equals $\frac{3\lambda}{4}$, as required.

Final Answer:
The second resonating length is 56.2 cm. This is option (D). \[ \boxed{\text{(D) }56.2\ \text{cm}} \]