JEE Advanced Model Paper 1 with solutions for online practice

Fundamental frequency of closed organ pipe is

v

4Ic

= 110 Hz ⇒ Ic =

330

4×110

= 0.75 m
First overtone of closed organ pipe =

3v

4Ic

First overtone of open organ pipe =

2v

2I0

=

v

I0

These two produce beats of frequency 2.2 Hz, when sounded together.
So,

3v

4Ic

-

v

I0

= 2.2 or

v

I0

-

3v

4lc

= 2.2

3

4×0.75

-

1

l0

=

2.2

330

⇒ I0 = 1.0067 m

1

I0

-

3

4×0.75

=

2.2

330

⇒ I0 = 0.9934 m