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Question

Three rods of same length and area of cross-section but of thermal conductivities K, 2K and 3K are connected in series in the same order. Free end of the first rod is at 0 C and free end of third rod is at 55 C. In steady state, temperature difference across the middle rod is

A
15 C
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B
30 C
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C
10 C
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D
25 C
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Solution

The correct option is A 15 C

Given that,
Let the length and area of each rod is l and A respectively and temperature of the junction is t1 and t2.
Rate of heat transfer by first rod = Rate of heat transfer by second rod = Rate of heat transfer by third rod.
H1=H2=H3
t10lAK=t2t1l2KA=55Ct2l3KA

3K(55t2)=2K(t2t1)=Kt1
3(55t2)=2(t2t1)=t1 (1)

From equation (1)
2t22t1=t1
t2=32t1 (2)

Again from equation (1)
3(55t2)=t1
1653t2=t1
165=t1+3t2
165=t1+(32×3)t1
330=11t1
t1=30 C
Now
t2t1=t12 (3)
t2t1=302=15 C
Hence, option (s) is correct.

(OR)

Given that,
Let the length and area of each rod is l and A respectively and temperature of the junction is t1 and t2.
Thermal resistance of first rod is R1=lKA
Thermal resistance of second rod is R2=l2KA
Thermal resistance of third rod is R3=l3KA

As these rods are in series heat flow through them can be considered as
H=ΔTR1+R2+R3=550lKA+l2KA+l3KA....i
Heat transfer through second rod is H=ΔT2R2=ΔT2l2KA....ii
Equating i & ii
550lKA+l2KA+l3KA=ΔT2l2KA

55011+12+13=ΔT212
ΔT2=55×311=15


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