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EAMCET Engineering 2004 Solved Paper
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© examsnet.com
Question : 21
Total: 160
The wavelengths of two notes in air are
36
195
m
and
36
193
m
. Each note produces 10 beats per second separately with a third note of fixed frequency. The velocity of sound in air in
m
∕
s
is
330
340
350
360
Validate
Solution:
Beat frequency
=
n
1
∼
n
2
Let the frequency of third note be
n
.
Then,
195
v
36
−
n
=
10
. . . (i)
and
n
−
193
v
36
=
10
. . . (ii)
Adding Eqs. (i) and (ii), we get
v
18
=
20
⇒
v
=
360
ms
−
1
© examsnet.com
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