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Two rods of same length and areas of cross-section A1 and A2 have their ends maintained at same temperature. If K1 and K2 are their thermal conductivities, C1 and C2 are the specific heats of their material and ρ1,ρ2 their densities, then for the rate of flow by conduction through them to be equal:

a
A1A2=K1c1ρ1c2ρ2K2
b
A1A2=K1K2
c
A1A2=K1c1ρ1K2c2ρ2
d
A1A2=K2K1

detailed solution

Correct option is D

∆Q∆t=K1A1(T1-T2)l=K2A2(T1-T2)l⇒A1A2=K2K1

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