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JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 1 Questions
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© examsnet.com
Question : 64
Total: 100
A large cylindrical rod of length L is made by joining two identical rods of copper and steel of length
(
l
2
)
each.
The rods are completely insulated from the surroundings. If the free end of copper rod is maintained at 100°C and that of steel at 0°C then the temperature of junction is (Thermal conductivity of copper is 9 times that of steel)
[Online May 19, 2012]
90°C
50°C
10°C
67°C
Validate
Solution:
Let conductivity of steel
K
steel
=
k
then from question
Conductivity of copper
K
copper
=
9
k
θ
copper
=
100
°
C
θ
steel
=
0
°
C
l
steel
=
l
copper
=
L
2
From formula temperature of junction;
θ
=
K
copper
θ
copper
l
steel
+
K
steel
θ
steel
l
copper
K
copper
l
steel
+
K
steel
l
copper
=
9
k
×
100
×
L
2
+
k
×
0
×
L
2
9
k
×
L
2
+
k
×
L
2
=
900
2
k
L
10
k
L
2
=
90
°
C
© examsnet.com
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