QJEEM23Phb37c: JEE Main 2023 Physics - Class 11 JEEM23Phb37c Solution: For a certain organ pipe, the first three resonance frequencies …

Given that the first three resonance frequencies are in the ratio of $1:3:5$ , we can express them as follows:

$f_1 = kf$ $f_3 = 3kf$ $f_5 = 5kf$

Where $k$ is a constant and $f_1, f_3$ , and $f_5$ are the first, third, and fifth resonance frequencies, respectively. We are given that the frequency of the fifth harmonic is $405 \mathrm{~Hz}$ , so we can write:

$f_5 = 5kf = 405 \mathrm{~Hz}$

Now we can solve for the constant $k$ :

$k = \frac{405}{5} = 81 \mathrm{~Hz}$

We also know that the speed of sound in air is $v = 324 \mathrm{~ms}^{-1}$ . The relationship between the speed of sound, the frequency, and the wavelength of a standing wave in a closed pipe can be expressed as follows:

$v = f\lambda$

Where $\lambda$ is the wavelength of the wave. For the first harmonic in a closed pipe, the length of the pipe is equal to one-fourth of the wavelength:

$L = \frac{1}{4}\lambda$

We can now substitute the expression for the wavelength in terms of the length into the equation for the speed of sound:

$v = f_1 \cdot 4L$

Now, we can substitute the value of $f_1 = kf = 81 \mathrm{~Hz}$ and the speed of sound $v = 324 \mathrm{~ms}^{-1}$ into the equation:

$324 = 81 \times 4L$

Now we can solve for the length of the organ pipe $L$ :

$L = \frac{324}{81 \times 4} = \frac{324}{324} = 1 \mathrm{~m}$

The length of the organ pipe is $1 \mathrm{~m}$ .