This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 10 Jan 2019 Shift 2)

Option 1 : 7

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
360 Marks
180 Mins

**Concept:**

Fundamental frequency of closed organ pipe is given by

\({f_0} = \frac{v}{{4L}}\)

where v is the velocity of sound in it and L is the length of the pipe.

Also, overtone frequencies are given by

\(f = \left( {2n + 1} \right)\frac{v}{{4L}}\)

Or

f = (2n+1) f_{0}

**Calculation:**

Given,

Fundamental frequency of the pipe f_{0} = 1.5 kHz = 1500 Hz

Highest frequency a person can hear, f_{max} = 20000 Hz

Given that, f_{0 }= 1500 Hz and f_{max} = 20000 Hz

This means, f_{max }> f

So, f_{max} > (2n + 1) f_{0}

20000 > (2n + 1)1500

2n + 1 < 13.33

2n < 13.33 – 1

2n < 12.33

n < 6.16

Or n = 6 (integer number)

Hence, a total of seven overtones will be heard (0 to 6)

Therefore, the number of overtones that can be heard by a person with this organ pipe will be seven (7).