MHTCET 23 Apr 2025 Shift 2 Paper

Concept:The overtone frequencies of open and closed organ pipes depend on their fundamental frequencies.Explanation:Let the fundamental frequency of the original open pipe be $f_o$.For an open pipe, the $n$th overtone has frequency $(n+1)f_o$.So the second overtone of the open pipe is $(2+1)f_o = 3f_o$.When the same pipe is closed at one end, its length $L$ does not change.The fundamental frequency of this closed pipe is $f_c = \frac{v}{4L} = \frac{f_o}{2}$.For a closed pipe, the $n$th overtone has frequency $(2n+1)f_c$.Thus, the third overtone of the closed pipe is $(2 \times 3 + 1)f_c = 7f_c = \frac{7f_o}{2}$.According to the question, this third overtone is $200\,\text{Hz}$ higher than the second overtone of the open pipe:$$\frac{7f_o}{2} = 3f_o + 200$$Multiplying both sides by 2 gives:$$7f_o = 6f_o + 400$$$$f_o = 400\,\text{Hz}$$Answer:The fundamental frequency of the open pipe is $400\,\text{Hz}$, which is Option C.