A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency that a person can hear is 20,000 Hz)

If a closed pipe vibration in $N^{th}$ mode then frequency of vibration 
 $n=\frac{(2N-1)v}{4l}=(2N-1)n_1$
(Where  $n_1=\text{fundamental frequency of vibration)}$
Hence  $20,000=(2N-1)\times 1500$   $\Rightarrow N=7.1\approx 7$

$$\begin{gathered} \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of over tones=(Number of mode of vibration)-1} \\ =7-1=6 \end{gathered}$$