MHTCET 19 Apr 2026 Shift 1 Paper

Concept:A closed pipe produces only odd harmonics, while an open pipe produces all harmonics.The third overtone corresponds to the $7^{\text{th}}$ harmonic for a closed pipe and the $4^{\text{th}}$ harmonic for an open pipe.Explanation:For a closed pipe of length $L_c$, the frequency of the $n^{\text{th}}$ harmonic is given by:$$f_c = \frac{(2n-1)v}{4L_c}$$The third overtone is the $7^{\text{th}}$ harmonic, so:$$f_c = \frac{7v}{4L_c}$$For an open pipe of length $L_0$, the frequency of the $n^{\text{th}}$ harmonic is given by:$$f_o = \frac{nv}{2L_0}$$The third overtone is the $4^{\text{th}}$ harmonic, so:$$f_o = \frac{4v}{2L_0} = \frac{2v}{L_0}$$Given that both frequencies are equal, $f_c = f_o$:$$\frac{7v}{4L_c} = \frac{2v}{L_0}$$Cancelling $v$ from both sides and cross-multiplying:$$7L_0 = 8L_c$$Therefore:$$\frac{L_c}{L_0} = \frac{7}{8}$$Hence, $L_c : L_0 = 7 : 8$.Answer:The required ratio is $7 : 8$, which corresponds to option B.