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AP EAMCET Engineering 2013 Solved Paper
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© examsnet.com
Question : 21
Total: 160
An air column in a tube 32 cm long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length 66 cm is in resonance with another tuning fork. When these two tuning forks are sounded together, they produce 8 beats per second. Then the frequencies of the two tuning forks are, (Consider fundamental frequencies only)
250 Hz, 258 Hz
240 Hz, 248 Hz
264 Hz, 256 Hz
280 Hz, 272 Hz
Validate
Solution:
We knows frequency of a closed end an column
n
1
=
V
4
I
1
We knows frequency of a open end an column
n
2
=
V
2
I
2
Given,
I
1
=
32
c
m
,
I
2
=
66
c
m
and
n
1
−
n
2
=
8
heat/s
So,
n
1
=
v
4
×
32
=
V
128
and
n
2
=
v
2
×
66
=
V
132
In given condition,
v
128
−
v
132
=
8
v
=
8448
×
4
v
=
33792
Hence,
n
1
=
33792
128
n
1
=
264
H
z
and
n
2
=
33792
132
n
2
=
256
H
z
© examsnet.com
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