Q 11-14-047JEE MainJEE Main 2024 (31 Jan, Shift 1)Easy
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is $60\ \text{cm}$, the length of the closed pipe will be:
Answer: (D) $15\ \text{cm}$
$$\frac{v}{4L_c} = \frac{2v}{2L_o} \Rightarrow L_c = \frac{L_o}{4} = \frac{60}{4} = 15\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics