A closed organ pipe and an open organ pipe are tuned to the same fundamental frequency. Determine the ratio of their lengths.
A. 1 : 1
B. 2 : 1
C. 1 : 4
D. 1 : 2

Hint: In this question, we need to find the ratio of closed and open organ pipes if they are tuned to the same frequency. So, we need to use the following formula. After, equating the equations for closed organ pipe and open organ pipe, we will get the desired result.

Formula used:
The formula for fundamental frequency for closed organ pipe is given by
\[{f_c} = \dfrac{v}{{4{L_c}}}\]
Similarly, the formula for fundamental frequency for open organ pipe is given by
\[{f_o} = \dfrac{v}{{2{L_o}}}\]
Where, \[{f_c}\] is the fundamental frequency for closed organ pipe, \[{f_o}\] is the fundamental frequency for open organ pipe, \[v\] is the speed of wave, \[{L_c}\] is the length of closed organ pipe and \[{L_o}\] is the length of open organ pipe.

Complete step by step solution:
We know that, the basic frequency for closed organ pipe is,
\[{f_c} = \dfrac{v}{{4{L_c}}}\]
Also, the basic frequency for closed organ pipe is,
\[{f_o} = \dfrac{v}{{2{L_o}}}\]
Here, speed is constant.
So, according to the given condition, both the pipes are tuned to the same frequency.Thus, we get
\[{f_c} = {f_o}\]
So, \[\dfrac{v}{{4{L_c}}} = \dfrac{v}{{2{L_o}}}\]
By simplifying, we get
\[\dfrac{1}{{4{L_c}}} = \dfrac{1}{{2{L_o}}}\]
\[\Rightarrow 4{L_c} = 2{L_o}\]
\[\Rightarrow 2{L_c} = {L_o}\]
By simplifying, further, we get
\[\dfrac{{{L_c}}}{{{L_o}}} = \dfrac{1}{2}\]
That means \[{L_c}:{L_o} = 1:2\]
Hence, the ratio of closed and open organ pipes, if they are tuned to the same frequency, is 1:2.

Therefore, the correct option is (D).

Note: Many students make mistakes in writing the formula for the fundamental frequency of a pipe. Consequently, the end result may get wrong. Here, the simplification part is also important for getting the final answer.