IIT JEE Advanced 2011 Paper 1

For a steady flow, we have

ΔQ

Δt

=

ΔT

RTh

where RTh is the thermal resistance, which is expressed as
RTh =

l

KA

If the width of each rod is taken as w, then we have
R1 =

L

2k(4Lw)

=

1

8kw

R2 =

4L

3k(Lw)

=

4

3kw

R3 =

4L

4k(2Lw)

=

1

2kw

R4 =

4L

5k(Lw)

=

4

5kw

R5 =

L

6k(4Lw)

=

1

24kw

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:
I1 = I5 = I
Therefore,
VPQ = IR1 ; VQR =

I

( 1 R2 )+( 1 R3 )+( 1 R4 )

; VRS = IRS
VPQ =

1

8kw

; VQR =

1

4kw

Therefore, the value of VRS is the least. That is,
VRS =

I

24kw

Now, I3 =

VQR

R3

=

I

4kw

×2kw =

I

2

I2 =

VQR

R2

=

I

4kw

×

3kw

4

=

3I

16

I4 =

VQR

R4

=

I

4kw

×

5kw

4

=

5I

16

Therefore,
I2+I4 = I2 = I3