Answer

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**Hint**

For solving this question, we need to find the fundamental frequency of the given open organ pipe. Then, with the help of the limit of the overtone frequencies, we can find the number of tones present.

Formula Used: The formula used to solve this question is given by

$ \nu = \dfrac{v}{{2l}} $, where $ \nu $ is the fundamental frequency of an organ pipe of length $ l $ and $ v $ is the value of the velocity of sound in the air.

**Complete step by step answer**

We know that the overtone frequencies in an organ pipe are the multiples of the fundamental frequency. The highest multiple of the fundamental frequency will be the last tone present in the open organ pipe. The number of the last tone will be the number of the tones present. So firstly we need to evaluate the value of the fundamental frequency.

We know that the fundamental frequency of an open organ pipe is given by

$ \nu = \dfrac{v}{{2l}} $

According to the question, we have $ v = 330m/s $ and $ l = 1m $. Putting these values in the above equation, we get

$ \nu = \dfrac{{330}}{{2 \times 1}} $

$ \nu = 165Hz $

Now, the overtone frequencies will be the multiples of this fundamental frequency. That is, the overtone frequencies will be

$ 330Hz,{\text{ }}495Hz,......... $

Now, limit of the frequencies present in the organ pipe is given as $ f \leqslant 1000 $

That is, the multiples of the fundamental frequency will not exceed $ 1000 $

For finding the multiple number of the highest frequency, we take the ratio of the highest frequency to the fundamental frequency.

$ ratio = \dfrac{{1000}}{{165}} = 6.06 $

As the overtone frequencies are the multiples of the fundamental frequency, so the number should be a natural number. So, taking the highest possible natural number less than or equal to the ratio, we get

$ n = 6 $

So, the $ 6th $ overtone will have the highest frequency in the given organ pipe.

Therefore there will be a total of $ 6 $ tones present.

**Hence, the correct answer is option (D).**

**Note**

Don’t get confused with the value of the upper limit of the frequency given in the question. It is just an estimate of the highest frequency of the overtones present in the organ pipe. It is not present in the organ pipe, but it gives an estimate of the number of tones.

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