An organ pipe is closed at one end and has fundamental frequency of $ 1500Hz $ . The maximum number of overtones generated by this pipe which a normal person can hear is
(A) 4
(B) 13
(C) 6
(D) 9

Hint : It is to be remembered that critical hearing frequency for a person is $ 20,000Hz $ . Only odd harmonics are produced in an organ pipe which closed at one end. So using arithmetic progression we can find the number of overtones from the range of $ 1500Hz $ to $ 20,000Hz $

Formula Used: The formulae used in the solution are given here.
 $\Rightarrow {n_1} + \left( {n - 1} \right)d \leqslant {\text{maximum upper limit}} $ where $ {n_1} $ is the first term and $ d $ is the difference between consequent terms of an arithmetic progression series.
 $\Rightarrow n = \dfrac{{\left( {2N - 1} \right)v}}{{4l}} = \left( {2N - 1} \right){n_1} $ where vibration is in the $ {N^{th}} $ mode and $ {n_1} $ is the fundamental frequency.

Complete step by step answer
An organ pipe is an instrument consisting of a closed-end column which typically contains a metal tube in which one of the ends is covered and not open to the surrounding air.
The natural frequency of the instrument is called harmonics and the harmonics in an organ pipe, closed at one end, is associated with standing patterns.
A closed-end instrument does not possess any even-numbered harmonics. Only odd-numbered harmonics are produced, where the frequency of each harmonic is some odd-numbered multiple of the frequency of the first harmonic. The next highest frequency above the third harmonic is the fifth harmonic.
Given that, the fundamental frequency of an organ pipe is closed at one end is $ 1500Hz $ .
It is known to us that the audible frequency of human ears is in between $ 20Hz $ to $ 20,000Hz $ .
So, the maximum frequency in the hearing range is $ 20,000Hz $ .
For $ f $ number of frequencies generated, the harmonics produced are in odd multiples of $ 1500Hz $ as $ {\text{1500, 4500, 7500,}}... $
As a result of this an arithmetic progression is formed.
Mathematically, for finding the $ {n^{th}} $ term of an arithmetic progression,
 $\Rightarrow 1500 + \left( {n - 1} \right) \times 3000 \leqslant 20000 $
On simplifying we get,
 $\Rightarrow 3000 \times n \leqslant 20000 + 1500 $
 $ \Rightarrow n \leqslant \dfrac{{21500}}{{3000}} $
The value of $ n $ is given by,
 $ \Rightarrow n = 7. $
So the number of audible overtones, is given by,
 $\Rightarrow \left( {n - 1} \right) = 6. $
 $ \therefore $ The number of overtones generated by this pipe which a normal person can hear is 6.
The correct answer is Option C.

Note
Alternatively we can also solve this problem as,
For a closed-end instrument,
 $\Rightarrow n = \dfrac{{\left( {2N - 1} \right)v}}{{4l}} = \left( {2N - 1} \right){n_1} $ where vibration is in the $ {N^{th}} $ mode and $ {n_1} $ is the fundamental frequency.
Assigning the values, $ n = 20000 $ and $ {n_1} = 1500 $ in the equation, we get,
 $\Rightarrow 20000 = \left( {2N - 1} \right) \times 1500 $
 $ \Rightarrow \dfrac{{20000}}{{1500}} = \left( {2N - 1} \right) $
Simplifying the equation further we have,
 $\Rightarrow \dfrac{{200}}{{15}} + 1 = 2N $
 $ \therefore $ The value of $ N $ is thus,
 $\Rightarrow \dfrac{{200 + 15}}{{15}} \times \dfrac{1}{2} = N $
 $ \Rightarrow N = 7.1 \approx 7. $
Thus, the harmonics produced are 1,3,5,7,9,11,13
The frequency of the $ {7^{th}} $ harmonic is $ 1500 \times 13 = 19500 $
This value is within the hearing range of humans.
The number of overtones heard is thus,
 $\Rightarrow N - 1 = 7 - 1 = 6 $
 $ \therefore $ The number of overtones generated by this pipe which a normal person can hear is 6. The correct answer is Option C.