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MHTCET 13 May 2023 Shift 2 Solved Paper
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Section:
Physics
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© examsnet.com
Question : 108
Total: 150
When two tuning forks are sounded together, 5 beats per second are heard. One of the forks is in unison with
0.97
m
length of sonometer wire and the other is in unison with
0.96
m
length of the same wire. The frequencies of the two tuning forks are
383
Hz
,
388
Hz
475
Hz
,
480
Hz
388
Hz
,
39
z
Hz
480
Hz
,
485
Hz
Validate
Solution:
Let,
f
1
and
f
2
be the frequency of the two turning forks, given
f
1
−
f
2
=
5
.
.
.
.
.
.
(i)
Length of wire in sonometer is
L
1
=
0.96
m
and
L
2
=
0.97
m
Frequency of vibration of the wire is given by
f
=
1
2
l
√
T
∕
m
According to the question,
f
1
f
2
=
l
2
l
1
=
0.97
0.96
>
1
∴
f
1
>
f
2
Now, from Eq. (i)
⇒
f
1
−
f
2
=
5
⇒
0.97
0.96
f
2
−
f
2
=
5
⇒
[
0.97
0.96
−
1
]
f
2
=
5
⇒
[
0.97
−
0.96
0.96
]
f
2
=
5
⇒
0.01
f
2
=
5
×
0.96
⇒
0.01
f
2
=
4.8
⇒
f
2
=
480
Hz
⇒
f
1
=
5
+
f
2
=
485
Hz
© examsnet.com
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