Resonance condition for a closed organ pipe / resonance column

For a resonance column, one end is closed by water and the other end is open. The effective length is: $L_{\text{eff}} = L + e$ where $e$ is the end correction.

Resonance occurs at: $L_{\text{eff}} = \frac{(2n-1)\lambda}{4}, \quad n=1,2,3,\dots$

Given data

• End correction: $e = 1\text{ cm}$
• Shortest resonating length: $L_1 = 10\text{ cm}$

For the shortest resonating length, this is the first resonance: $L_1 + e = \frac{\lambda}{4}$

So, $10 + 1 = \frac{\lambda}{4}$ $11 = \frac{\lambda}{4}$ $\lambda = 44\text{ cm}$

Next resonating length

The next resonance in a closed pipe is: $L_2 + e = \frac{3\lambda}{4}$

Substitute $\lambda = 44$ cm: $L_2 + 1 = \frac{3\times 44}{4} = 33$ $L_2 = 33 - 1 = 32\text{ cm}$

Check using spacing of resonances

Successive resonating lengths differ by: $\frac{\lambda}{2} = \frac{44}{2} = 22\text{ cm}$

Hence, $L_2 = 10 + 22 = 32\text{ cm}$

This confirms the result.

Option matching

$\boxed{32\text{ cm}}$ So the correct option is B.