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IIT JEE Advanced 2011 Paper 1
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© examsnet.com
Question : 10
Total: 69
A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then, in steady-state
[JEE Adv 2011 P1]
heat flow through A and E slabs are same.
heat flow through slab E is maximum.
temperature difference across slab E is smallest.
heat flow through C = heat flow through B + heat flow through D.
Validate
Solution:
For a steady flow, we have
Δ
Q
Δ
t
=
Δ
T
R
T
h
where
R
T
h
is the thermal resistance, which is expressed as
R
T
h
=
l
K
A
If the width of each rod is taken as w, then we have
R
1
=
L
2
k
(
4
L
w
)
=
1
8
k
w
R
2
=
4
L
3
k
(
L
w
)
=
4
3
k
w
R
3
=
4
L
4
k
(
2
L
w
)
=
1
2
k
w
R
4
=
4
L
5
k
(
L
w
)
=
4
5
k
w
R
5
=
L
6
k
(
4
L
w
)
=
1
24
k
w
The equivalent electrical circuit for the above set-up can be depicted as shown in the following figure:
The heat flow through each rod is equivalent to current through each thermal resistance:
I
1
=
I
5
= I
Therefore,
V
P
Q
=
I
R
1
;
V
Q
R
=
I
(
1
R
2
)
+
(
1
R
3
)
+
(
1
R
4
)
;
V
R
S
=
I
R
S
V
P
Q
=
1
8
k
w
;
V
Q
R
=
1
4
k
w
Therefore, the value of
V
R
S
is the least. That is,
V
R
S
=
I
24
k
w
Now,
I
3
=
V
Q
R
R
3
=
I
4
k
w
×
2
k
w
=
I
2
I
2
=
V
Q
R
R
2
=
I
4
k
w
×
3
k
w
4
=
3
I
16
I
4
=
V
Q
R
R
4
=
I
4
k
w
×
5
k
w
4
=
5
I
16
Therefore,
I
2
+
I
4
=
I
2
=
I
3
© examsnet.com
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