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IIT JEE Advanced 2011 Paper 1
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Section:
Physics
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© examsnet.com
Question : 19
Total: 69
Steel wire of length L at 40°C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length L. The coefficient of linear thermal expansion of the steel is
10
−
5
∕
°
C
, Young’s modulus of steel is
10
11
N
∕
m
2
and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of m in kg is nearly ______.
[JEE Adv 2011 P1]
Your Answer:
Validate
Solution:
The change in length is
ΔL =
F
L
Y
A
F = mg
The change in length due to the change in temperature is ΔL = LαΔT
where α is the coefficient of linear expansion and ΔT is the change in temperature.
α =
10
−
5
∕
°
C
, Y =
10
11
N
m
2
The radius of wire is 1 mm. Therefore,
A =
π
×
10
−
6
m
2
Now,
F
L
Y
A
= LαΔT
That is, mg = YAαΔT
Therefore,
m =
0
11
×
3.14
×
10
−
6
×
10
−
5
×
10
9.8
= 3.2 kg ~ 3 kg
© examsnet.com
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