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JEE Advanced Model Paper 4 with solutions for online practice
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© examsnet.com
Question : 14
Total: 54
Two plates of thickness
l
1
and
l
2
with thermal conductivities
K
1
and
K
2
respectively are kept in thermal contact as shown in the figure
If' A' is area of each face and beat is flowing through the combination in steadystate along the direction as shown in the figure. Then mark the correct options.
thermal resistance of combination is R =
l
1
A
K
1
+
l
2
A
K
2
Heat flow rate through the combination is H =
A
(
T
1
−
T
2
)
(
l
1
K
1
+
l
2
K
2
)
Temperature of joint of slabs is T =
T
1
l
2
K
2
+
l
1
K
1
T
2
(
l
1
K
1
+
l
2
K
2
)
Equivalent thermal conductivity of combination will be
k
e
q
=
l
1
+
l
2
(
l
1
K
1
+
l
2
K
2
)
Validate
Solution:
R
e
q
=
R
1
+
R
2
=
l
1
A
K
1
+
l
2
A
K
2
H =
d
Q
d
t
=
A
Δ
θ
R
=
(
T
1
−
T
2
)
A
R
1
+
R
2
-
(
T
1
−
T
2
)
A
(
l
1
K
1
+
l
2
K
2
)
Also ,
T
1
−
T
2
R
1
+
R
2
=
T
−
T
2
R
1
⇒ T =
T
1
l
2
K
2
+
T
2
l
1
K
1
(
l
1
K
1
+
l
2
K
2
)
And R =
R
1
+
R
2
So,
K
e
q
=
l
1
+
l
2
A
(
R
1
+
R
2
)
=
l
1
+
l
2
(
l
1
K
1
+
l
2
K
2
)
© examsnet.com
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