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Question

The three rods of lenght L and cross-sectional area A is described in the above problem are placed individually, with their ends kept at the same temperature difference. The rate of heat flow through C is equal to the rate of combined heat flow through A and B kc must be equal to
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A
kA+kB
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B
kAkBkA+kB
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C
12(kA+kB)
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D
2(kAkBkA+kB)
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Solution

The correct option is A kA+kB
As we know that Qt=KA(TATB)L,and all rods are kept separately,so we write individual equation for all and apply the given relation between them.
KCA(T1T2)L=KAA(T1T2)L=KBA(T1T2)L
KC=KA+KB

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