Step 1: Identify the type of tube and the wave pattern. The diagram shows a tube that is closed at one end (bottom, indicated by shading) and open at the other end (top). For a stationary wave in a closed tube, there must be a node (point of zero displacement) at the closed end and an antinode (point of maximum displacement) at the open end. Step 2: Determine the relationship between the length of the air column ($L$) and the wavelength ($\lambda$) from the diagram. The diagram shows the following pattern from the closed end to the open end: A node at the closed end. An antinode. Another node. An antinode at the open end. The distance from a node to an adjacent antinode is $\frac{\lambda}{4}$. The distance from an antinode to an adjacent node is also $\frac{\lambda}{4}$. Tracing the wave pattern in the diagram: From the node at the closed end to the first antinode: $\frac{\lambda}{4}$ From the first antinode to the next node: $\frac{\lambda}{4}$ From the second node to the antinode at the open end: $\frac{\lambda}{4}$ Therefore, the total length $L$ of the resonating air column is the sum of these segments: $$L = \frac{\lambda}{4} + \frac{\lambda}{4} + \frac{\lambda}{4} = \frac{3\lambda}{4}$$ This corresponds to the first overtone (or third harmonic) for a closed tube. Step 3: Substitute the given wavelength and calculate $L$. The wavelength is given as $\lambda = 40 \, cm$. $$L = \frac{3 \times 40 \, cm}{4}$$ $$L = 3 \times 10 \, cm$$ $$L = 30 \, cm$$ Step 4: Compare the result with the given options. The calculated length $L = 30 \, cm$ matches option (d). The final answer is $\boxed{\text{D. 30cm}}$.