An open pipe 30 cm long and closed pipe 23 cm long, both of the same diameter, are each sounding its first overtone and these are in unison. The end correction of these pipes is:
A. 0.5cm
B. 0.3cm
C. 1cm
D. 1.2cm

Hint:-The concept to be applied here, to solve this problem is that of standing waves.
If the wave travels to one end and gets reflected, and this reflected wave in turn, gets reflected again, setting up a steady wave pattern in the medium, these waves are called standing waves.
The standing waves have different sets of frequencies and the smallest frequency is called the fundamental frequency.

Complete step-by-step solution:-
If the wave travels to one end and gets reflected, and this reflected wave in turn, gets reflected again, setting up a steady wave pattern in the medium, these waves are called standing waves. There can be several values of amplitude across the standing wave but the points at which the amplitude is zero are called the nodes.
The difference between the nodes is equal to $\dfrac{\lambda }{2}$ where $\lambda $ is the wavelength.
If we consider pipe of air column of length L, we have the relation among the wavelength, velocity and the length as follows:
i) For pipe open on both ends:
Frequency, $\upsilon = n\dfrac{v}{{2L}}$
i) For pipe closed at one end:
Frequency, $\upsilon = \left( {n + \dfrac{1}{2}} \right)\dfrac{v}{{2L}}$
As you can see, in the open pipe, we have even multiples of $\dfrac{v}{{2L}}$ i.e. $0$, $\dfrac{v}{L}$, $2\dfrac{v}{L}$, $3\dfrac{v}{L}$
In the closed pipe, we have odd multiples of $\dfrac{v}{{2L}}$ i.e. $\dfrac{v}{{4L}}$, $3\dfrac{v}{{4L}}$, $5\dfrac{v}{{4L}}$, $7\dfrac{v}{{4L}}$
The first frequency where n = 0 is called the fundamental frequencies and the subsequent frequencies are called overtones or harmonics such as first overtone/harmonic, second overtone/harmonic, etc.
Thus, the first overtones of the open and closed pipe are given as follows:
Open pipe, ${\upsilon _1} = \dfrac{v}{{{L_o}}}$
Closed pipe, ${\upsilon _1} = 3\dfrac{v}{{4{L_c}}}$
Since they are equal as per the problem, we have –
$\dfrac{v}{{{L_o}}} = \dfrac{{3v}}{{4{L_c}}}$
Thus, $\dfrac{{{L_c}}}{{{L_o}}} = \dfrac{3}{4}$
However, the students must note that these lengths are not exactly equal to the lengths of the pipes because the node and antinode formation in them occur at a small distance away from the ends. Thus, a small value of length called end correction, denoted by $e$, should be added to the actual lengths of the pipes to make it equal to the above formula.
The formula for end correction is,
Open pipe – ${L_o} = L + 2e$ where L is the length of the open pipe, L = 30cm
Closed pipe – ${L_o} = L + e$ where L is the length of closed pipe, L = 23cm
Substituting them in the above equation, we get –
$\Rightarrow 4\left( 23+e \right)=3\left( 30+2e \right)$
$\Rightarrow 92+4e=90+6e$
$\Rightarrow 6e-4e=92-90=2$
$\Rightarrow 2e=2$
Therefore, end correction, $e$= 1 cm
Hence, the correct option is Option C.

Note:- The end correction is also, approximately equal to 60% of the radius of the pipe.
$e = 0.6r$. So, sometimes, in a similar question, they might give you the radii of the open and closed pipes. Then, instead of their lengths, you can use the above relation to find the end corrections.