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Question

Two conducting rods A and B of same length and cross-sectional area are connected (i) In series (ii) In parallel as shown. In both combination a temperature difference of 100 C​ is maintained. If thermal conductivity of A is 3K and that of B is K then the ratio of heat current flowing in parallel combination to that flowing in series combination is


A

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B
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C
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D
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Solution

The correct option is A

Heat current H=ΔθRHpHs=RsRp
In first case : Rs=R1+R2=l(3K)A+lKA=43lKA
In second case : Rp=R1R2R1+R2=1(3K)A×lKA(l(3K)A+lKA)=l4KA
HpHs=4l3KAl4KA=163


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