JEE Physics 2024 Question with Solution

Given: The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. The length of the open pipe is $60 \, \text{cm}$.

Find: The length of the closed pipe.

For a closed organ pipe, the fundamental frequency is

$$f_{\text{closed}} = \frac{v}{4L_{\text{closed}}}$$

For an open organ pipe, the first overtone is the second harmonic, so its frequency is

$$f_{\text{open}} = \frac{v}{L_{\text{open}}}$$

According to the question,

$$f_{\text{closed}} = f_{\text{open}}$$

Therefore,

$$\frac{v}{4L_{\text{closed}}} = \frac{v}{L_{\text{open}}}$$

Cancelling $v$ from both sides,

$$4L_{\text{closed}} = L_{\text{open}}$$

So,

$$L_{\text{closed}} = \frac{L_{\text{open}}}{4}$$

Substituting $L_{\text{open}} = 60 \, \text{cm}$,

$$L_{\text{closed}} = \frac{60}{4} = 15 \, \text{cm}$$

Therefore, the length of the closed pipe is $15 \, \text{cm}$. The correct option is D.

Given: The fundamental frequency of a closed organ pipe equals the first overtone frequency of an open organ pipe. Let the closed pipe length be $L_1$ and the open pipe length be $L_2 = 60 \, \text{cm}$.

Find: $L_1$.

For the fundamental mode of a closed organ pipe,

$$L_1 = \frac{\lambda}{4}$$

For the first overtone of an open organ pipe, the pipe contains one full wavelength, so

$$L_2 = \lambda$$

Using $v = f\lambda$ and the fact that the frequencies are equal, both pipes correspond to the same wavelength relation for the stated modes. Hence,

$$L_2 = 4L_1$$

Now substitute $L_2 = 60 \, \text{cm}$,

$$60 = 4L_1$$ $$L_1 = 15 \, \text{cm}$$

Therefore, the length of the closed pipe is $15 \, \text{cm}$, so the correct option is D.