The end correction of a resonance column is 1 cm. If the shortest length resonating with the tuning fork is 10 cm, the next resonating length should be:

For a resonance column that is closed at one end and open at the other, stationary (standing) waves form only at specific lengths. The condition for resonance is

$$\text{(effective length)} = \left(n+\dfrac12\right)\dfrac{\lambda}{2}, \qquad n = 0,1,2,\ldots$$

Because the tube is closed at one end, the fundamental (shortest) resonance corresponds to a node at the closed end and an antinode just outside the open end. Experimentally, the air antinode is not exactly at the rim; it is slightly outside. This shift is called the end correction and is denoted by $$e$$. Therefore

$$\text{effective length} = (\text{length of air column}) + e.$$

Let the wavelength of the sound produced by the tuning fork be $$\lambda$$. For the shortest resonance we have

$$L_1 + e = \dfrac{\lambda}{4}$$ (because the effective length equals one-quarter wavelength).

We are given

$$L_1 = 10\ \text{cm}, \qquad e = 1\ \text{cm}.$$

Substituting these values, we calculate the wavelength:

$$10 + 1 = \dfrac{\lambda}{4} \;\Longrightarrow\; 11 = \dfrac{\lambda}{4} \;\Longrightarrow\; \lambda = 4 \times 11 = 44\ \text{cm}.$$

The next resonance in a closed tube occurs when the effective length equals $$\dfrac{3\lambda}{4}$$ (corresponding to the third harmonic, with two additional nodes and antinodes inside the tube). Hence we write

$$L_2 + e = \dfrac{3\lambda}{4}.$$

Substituting $$\lambda = 44\ \text{cm}$$ and $$e = 1\ \text{cm}$$, we obtain

$$L_2 + 1 = \dfrac{3 \times 44}{4} = \dfrac{132}{4} = 33\ \text{cm}.$$

Now isolate $$L_2$$:

$$L_2 = 33 - 1 = 32\ \text{cm}.$$

This length is the next (second) resonating length of the air column.

Hence, the correct answer is Option A.