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JEE Main 26 Feb 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 2
Total: 89
The length of metallic wire is
I
1
when tension in it is
T
1
. It is
I
2
when the tension is
T
2
. The original length of the wire will be
[26 Feb 2021 Shift 2]
I
1
+
I
2
2
T
2
I
1
+
T
1
I
2
T
1
+
T
2
T
2
I
1
−
T
1
I
2
T
2
−
T
1
T
1
l
1
−
T
2
I
2
T
2
−
T
1
Validate
Solution:
Let
I
0
be the original length,
A
be the area of cross-section,
α
be the coefficient of linear expansion,
∆
l
be the change in length and
Y
be the Young's modulus of elasticity.
As,
I
1
=
I
0
(
1
+
α
∆
T
)
⇒
I
1
−
I
0
=
I
0
α
∆
T
⇒
∆
I
=
I
0
α
∆
T
Initially,
Y
=
Stress
Strain
=
T
∕
A
∆
l
∕
I
0
⇒
Y
=
T
1
∕
A
(
I
1
−
I
0
)
∕
I
0
. . . (i)
Finally,
Y
=
T
2
∕
A
(
I
2
−
I
0
)
∕
I
0
. . . (ii)
Now, from Eqs. (i) and (ii), we get
T
1
∕
A
(
I
1
−
I
0
)
∕
I
0
=
T
2
∕
A
(
I
2
−
I
0
)
∕
I
0
⇒
T
1
I
1
−
I
0
=
T
2
I
2
−
I
0
⇒
T
1
I
2
−
T
1
I
0
=
T
2
I
1
−
T
2
I
0
⇒
T
1
I
2
−
T
2
I
1
=
(
T
1
−
T
2
)
I
0
⇒
I
0
=
T
1
I
2
−
T
2
I
1
T
1
−
T
2
or
I
0
=
T
2
I
1
−
T
1
I
2
T
2
−
T
1
© examsnet.com
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