Q.

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  4 cals-1, in shorter it will be  (in cals-1)

Moderate

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By Expert Faculty of Sri Chaitanya

a

1

b

2

c

3

d

8

answer is 1.

(Detailed Solution Below)

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Detailed Solution

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
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