Heat is flowing in steady state from temperature 100oC to temperature 0oC through a rod of length l, cross sectional area A and thermal conductivity K at a rate Q. When another rod of length 2l, cross-sectional area A and thermal conductivity K/2 is joined end to end to the above rod to make a composite rod of length 3l whose ends are maintained at 100oC and 0oC, rate of heat conduction at steady state is
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a
2Q5
b
Q5
c
3Q5
d
4Q5
answer is B.
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Detailed Solution
For the first rod, thermal resistance R=lKA=100-0QFor the second rod, thermal resistance R'=2lK/2A=4lKA=4RSo equivalent thermal resistance Re=R+R'=R+4R=5R∴Q'=100-0Re=1005R=15 x Q