MHTCET 15 Apr 2026 Shift 2 Paper

Concept:For an open pipe, harmonics are $1f, 2f, 3f, \dots$; for a closed pipe, only odd harmonics $f_c, 3f_c, 5f_c, \dots$ are present.Explanation:Let the fundamental frequency of the open pipe be $f = \dfrac{v}{2L}$.The first overtone of the open pipe is its second harmonic:$$f_{\text{open, 1st overtone}} = 2f$$When the same pipe is closed at one end, its fundamental frequency becomes:$$f_c = \dfrac{v}{4L} = \dfrac{1}{2}\cdot\dfrac{v}{2L} = \dfrac{f}{2}$$For a closed pipe, the second overtone is the fifth harmonic:$$f_{\text{closed, 2nd overtone}} = 5f_c = 5\left(\dfrac{f}{2}\right) = \dfrac{5f}{2}$$Given that this is $100\ \text{Hz}$ higher than the first overtone of the open pipe:$$\dfrac{5f}{2} - 2f = 100$$$$\dfrac{f}{2} = 100$$$$f = 200\ \text{Hz}$$Answer:The fundamental frequency of the open pipe is $200\ \text{Hz}$.Correct option: A. $200\ \text{Hz}$