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# The ends of the two rods of conductivities, radii and lengths in the ratio of 1:2 are maintained at the same temperature difference, if the rate of flow of heat through the bigger rod  is  , in shorter it will be

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a
1
b
2
c
3
d
8
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detailed solution

Correct option is A

Rate of flow of heat is P P=KA(θ1-θ2)l=Kπr2(θ1-θ2)l here K is coefficient of thermal conductivity, A is area of cross section, θ1,θ2 temperature at each end of rod l is length of rod substitute given values legibly  P1P2=K1K2.r1r22×l2l1=1 2122x21 given P2=4 calorie second-1 P4=14 rate of flow of heat through short rod is P=1 cals-1

Similar Questions

Two rods of length ${\mathrm{d}}_{1}$ and ${\mathrm{d}}_{2}$ and coefficients of thermal conductivities ${\mathrm{K}}_{1}$ and ${\mathrm{K}}_{2}$ are kept touching each other. Both have the same area of cross-section. The equivalent of thermal conductivity is

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