An open organ pipe has a fundamental frequency of 300Hz. The first overtone of a closed pipe has the same frequency as the first overtone of an open organ pipe. How long is the closed pipe?
A. \[41.25\,{\text{cm}}\]
B. \[42.3\,{\text{cm}}\]
C. \[49.5\,{\text{cm}}\]
D. \[40.5\,{\text{cm}}\]

Hint: First of all, we will use the expression of the fundamental frequency of the open pipe to find the length of the pipe. Then we will again use the formula of frequency of the first overtone of a closed pipe. We will equate this expression with the expression of frequency of the first overtone of the open pipe. We will manipulate accordingly and obtain the result.

Complete step by step solution:
In the given question, we are supplied the following data:
The fundamental frequency of an open organ pipe is given as \[300\,{\text{Hz}}\] .We are also told that the first overtone of a closed pipe has the same frequency as the first overtone of open organ pipe.We are asked to determine the length of the closed pipe.

To begin with, we will first analyse the situation as given in the question. First, we will try to find the length of the open pipe using the formula which relates fundamental frequency, velocity of the sound and the length of the pipe. After that we will go for the second pipe that is the closed pipe.

Let us proceed to solve the numerical.
For this we will take the velocity of sound in air to be \[330\,{\text{m}}\,{{\text{s}}^{ - 1}}\] .
First of all, we will work on the open pipe.
The formula which relates fundamental frequency, velocity of the sound and the length of the pipe is given below:
\[{n_0} = \dfrac{v}{{2l}}\] …… (1)
Where,
\[{n_0}\] indicates the fundamental frequency of the open organ pipe.
\[v\] indicates the velocity of sound in air.
\[l\] indicates the length of the organ pipe.

Now, we will substitute the required values in the equation (1) and we get:
${n_0} = \dfrac{v}{{2l}} \\
\Rightarrow l = \dfrac{v}{{2{n_0}}} \\
\Rightarrow l = \dfrac{{330}}{{2 \times 300}} \\
\Rightarrow l = \dfrac{{11}}{{20}}\,{\text{m}}$
Therefore, the length of the open organ pipe is found to be \[\dfrac{{11}}{{20}}\,{\text{m}}\] .
Again, we know that the frequency of the first overtone of a closed pipe is given as:
\[n = 3 \times \dfrac{v}{{4l'}}\] …… (2)
Where,
\[n\] indicates the frequency of the first overtone of the closed pipe.
\[v\] indicates the velocity of sound in air.
\[l'\] indicates the length of the closed pipe.

We know the frequency of the first overtone of an open pipe is given as:
\[n' = 2 \times \dfrac{v}{{2l}}\]
Where,
\[n'\] indicates the frequency of the first overtone of the open pipe.
\[v\] indicates the velocity of sound in air.
\[l\] indicates the length of the closed pipe.

The frequency of the first overtone of an open pipe is equal to the frequency of first overtone of a closed pipe.
Then we can write:
$n' = n \\
\Rightarrow 2 \times \dfrac{v}{{2l}} = 3 \times \dfrac{v}{{4l'}} \\
\Rightarrow \dfrac{v}{l} = \dfrac{{3 \times v}}{{4l'}} \\
\Rightarrow l' = \dfrac{{3 \times l}}{4} \\
\Rightarrow l' = \dfrac{3}{4} \times \dfrac{{11}}{{20}} \\
\Rightarrow l' = 0.4125\,{\text{m}} \\
\therefore l' = 41.25\,{\text{cm}}$

Hence, the length of the closed pipe is \[41.25\,{\text{cm}}\] .The correct option is A.

Note: While solving the problem most of the students tend to make mistakes in finding the correct expressions for the frequency for the first overtone. It is important to remember that in the case of the closed pipe, the first harmonic is represented by a wavelength which is four times the length of the pipe. The frequency of the first overtone is three times the first harmonic.