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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper
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© examsnet.com
Question : 99
Total: 160
At a temperature of 314 K and a pressure of 100 kPa. the speed of sound in a gas is 1380
m
s
−
1
. The radius of each gas molecule is 0.5 A. The frequency of sound at which the wavelength of sound wave in the gas becomes equal to the mean free path of the gas molecules is
(Boltzmann Constant = 1.38 x
10
−
23
J
K
−
1
)
1000 MHz
1000
√
2
MHz
1000
√
2
MHz
500 MHz
Validate
Solution:
The expression for the mean free path is given by,
λ
=
k
T
√
2
π
d
2
p
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
(
I
)
The expression for the frequency is given by,
v
=
v
λ
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
(
I
I
)
From equation (I) and (II),
v
=
√
2
π
d
2
p
×
v
k
T
Substitute the given values in the above equation.
v
=
√
2
(
3.14
)
(
10
−
20
)
(
10
5
)
×
1380
(
1.38
×
10
−
23
)
×
314
=
√
2
×
10
9
H
z
=
1000
√
2
M
H
z
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