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AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper
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© examsnet.com
Question : 93
Total: 160
When the area of cross section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes
k
times the initial speed. Then
k
=
1
4
2
8
Validate
Solution:
Speed of transverse wave in stretched wire is given as
v
=
√
T
m
where,
T
=
tension in the wire
and
m
=
mass per unit length of wire.
∴
v
=
√
T
M
∕
l
[
∵
m
=
M
l
]
=
√
T
M
=
√
T
V
⋅
ρ
[
∵
M
=
V
⋅
ρ
]
=
√
T
l
A
⋅
l
⋅
0
[
∵
V
=
A
⋅
l
]
v
=
√
T
A
ρ
⇒
v
∝
√
T
A
⇒
v
2
v
1
=
√
T
2
T
1
×
A
1
A
2
...(i)
Given,
A
2
=
A
1
2
,
T
2
=
2
T
1
∴
From Eq. (i), we get
v
2
v
1
=
√
2
T
1
T
1
×
A
1
A
1
=
√
4
v
2
=
2
v
1
⇒
K
=
2
© examsnet.com
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