MHTCET 11 Apr 2026 Shift 2 Paper

Concept:For an open pipe, the first overtone frequency is $f_o = \frac{v}{L}$. For a closed pipe, the first overtone frequency is $f_c = \frac{3v}{4L}$. After changing the lengths, use the fundamental frequency of each pipe.Explanation:Let the original length of each pipe be $L$ and speed of sound be $v$.Initial first overtone frequencies:$f_o = \frac{v}{L}$, $f_c = \frac{3v}{4L}$.They produce 3 beats, so:$f_o - f_c = 3$$\frac{v}{L} - \frac{3v}{4L} = 3$$\frac{v}{4L} = 3 \Rightarrow \frac{v}{L} = 12$.New length of open pipe $= \frac{L}{3}$.Its new fundamental frequency:$f_o' = \frac{v}{2(L/3)} = \frac{3v}{2L} = \frac{3}{2} \times 12 = 18$ Hz.New length of closed pipe $= 3L$.Its new fundamental frequency:$f_c' = \frac{v}{4(3L)} = \frac{v}{12L} = \frac{1}{12} \times 12 = 1$ Hz.New beat frequency:$|18 - 1| = 17$ beats.Answer:$17$ beats, i.e., Option C.