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JEE Advanced 2020 Full Test 4 Paper 1
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© examsnet.com
Question : 8
Total: 54
A steel wire of length
60
c
m
and area of cross
A
section
10
−
6
m
2
is joined with an aluminium wire of length
45
c
m
and area of cross section
3
×
10
−
6
m
2
. The composite string is stretched by a tension of 80 N. Density of steel is
7800
k
g
m
−
3
and that of aluminium is
2600
k
g
m
−
3
. The minimum frequency of turning fork, which can produce standing wave in it with node at joint is
357.3
H
z
375.3
H
z
337.5
H
z
325.3
H
z
Validate
Solution:
Mass per unit length,
µ
=
m
ℓ
=
S
A
ℓ
ℓ
=
ρ
A
µ
s
=
µ
A
ℓ
=
78
×
10
−
4
k
g
∕
m
∴
Speed of wave is same in both wire
V
=
√
T
µ
=
√
80
×
10
4
78
=
2
×
10
2
√
3.9
v
min
=
V
λ
max
=
200
√
3.9
×
0.3
[
λ
max
2
=
15
c
m
for
C
as a node
]
=
337.4
H
2
© examsnet.com
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