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VITEEE 2018 Solved Paper
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© examsnet.com
Question : 20
Total: 125
The length of elastic string, obeying Hooke’s law is
l
1
metres when the tension 4N and
l
2
metres when the tension is 5N. The length in metres when the tension is 9N is -
5
l
1
−
4
l
2
5
l
2
−
4
l
1
9
l
1
−
8
l
2
9
l
2
−
8
l
1
Validate
Solution:
Let
ℓ
0
be the unstretched length and
ℓ
3
be the length under a tension of
9
N
.
Then
Y
=
4
ℓ
0
A
(
ℓ
1
−
ℓ
0
)
=
5
ℓ
0
A
(
ℓ
2
−
ℓ
0
)
=
9
ℓ
0
A
(
ℓ
3
−
ℓ
0
)
These give
4
ℓ
1
−
ℓ
0
=
5
ℓ
2
−
ℓ
0
⇒
ℓ
0
=
5
ℓ
1
−
4
ℓ
2
Further,
4
ℓ
1
−
ℓ
0
=
9
ℓ
2
−
ℓ
0
Substituting the value of
ℓ
0
and solving,
we get
ℓ
3
=
5
ℓ
2
−
4
ℓ
1
© examsnet.com
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