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AP EAMCET Engineering 2011 Solved Paper
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© examsnet.com
Question : 25
Total: 160
The wavelengths of two sound notes in air are
40
195
m
and
40
193
m
.
Each note produces 9 beats per second separately with a third note of fixed frequency. The velocity of sound in air in
m
∕
s
is
360
320
300
340
Validate
Solution:
Here,
v
1
=
v
(
40
195
)
and
v
2
=
v
(
40
193
)
⇒
v
1
=
v
(
195
)
40
and
v
2
=
v
(
193
)
40
According to given condition
v
40
(
195
)
−
v
=
9
.......(i)
and
v
−
v
(
193
)
40
=
9
.......(ii)
Adding Eqs. (i) and (ii)
18
=
v
40
(
195
−
193
)
=
2
v
40
⇒
18
=
v
20
or
v
=
360
m
∕
s
© examsnet.com
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