For sound waves, if the number of nodes for the $5^{\text{th}}$ harmonic of an open-ended pipe is $n$ and that for the $9^{\text{th}}$ harmonic of the same pipe with one of its ends closed is $m$, the ratio $\dfrac{n}{m}$ is :
Answer: (C) $1$
Concept. Count nodes from the standing-wave pattern of each pipe: an open pipe has antinodes at both ends, a closed pipe has a node at the closed end and an antinode at the open end.
Step 1. Open pipe, $5^{\text{th}}$ harmonic. $L = 5\dfrac{\lambda}{2}$: five half-loops, each with one node inside, so $n = 5$.
Step 2. Closed pipe, $9^{\text{th}}$ harmonic. $L = 9\dfrac{\lambda}{4}$, i.e. $(2k - 1) = 9$ with $k = 5$. The pattern has $k = 5$ nodes (counting the closed end), so $m = 5$.
Step 3. Ratio. $\dfrac{n}{m} = \dfrac55 = 1$.
Answer. $1$ (Option C).
Solution by Sreeraj P, M.Sc Physics