MHTCET 16 Apr 2026 Shift 2 Paper

Concept:For an open pipe, all harmonics are present.For a closed pipe, only odd harmonics are present.Explanation:Let the fundamental frequency of the open pipe be $f_o$.For an open pipe of length $L$,$$f_o = \frac{v}{2L}$$The second overtone of an open pipe is the third harmonic.So its frequency is$$3f_o$$When the same pipe is closed at one end, its fundamental frequency becomes$$f_c = \frac{v}{4L} = \frac{f_o}{2}$$The third overtone of a closed pipe is the $7$th harmonic.Thus, its frequency is$$7f_c = 7\left(\frac{f_o}{2}\right) = \frac{7f_o}{2}$$This is given to be $150\ \mathrm{Hz}$ higher than the second overtone of the open pipe.So,$$\frac{7f_o}{2} - 3f_o = 150$$$$\frac{7f_o - 6f_o}{2} = 150$$$$\frac{f_o}{2} = 150$$$$f_o = 300\ \mathrm{Hz}$$Answer:The fundamental frequency of the open pipe is $300\ \mathrm{Hz}$.Correct option: D.