MHTCET 5 May 2025 Shift 2 Paper

Concept:Beats are produced due to the difference in the fundamental frequencies of the two pipes.Explanation:For an open pipe, fundamental frequency is $n_o = \frac{v}{2L}$.For a closed pipe, fundamental frequency is $n_c = \frac{v}{4L}$.Given initial beats: $n_o - n_c = 2$.So, $\frac{v}{2L} - \frac{v}{4L} = 2$.This simplifies to $\frac{v}{4L} = 2$, hence $\frac{v}{L} = 8$.Now, the open pipe length is halved, so $L_o' = \frac{L}{2}$.New open pipe frequency: $n_o' = \frac{v}{2\left(\frac{L}{2}\right)} = \frac{v}{L}$.The closed pipe length is doubled, so $L_c' = 2L$.New closed pipe frequency: $n_c' = \frac{v}{4(2L)} = \frac{v}{8L}$.New beat frequency: $n_o' - n_c' = \frac{v}{L} - \frac{v}{8L} = \frac{7v}{8L}$.Using $\frac{v}{L} = 8$: $\frac{7v}{8L} = \frac{7}{8} \times 8 = 7$.Answer:$7$ beats per second (Option C).