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Two sheets of thickness d and 3d, are to...

Two sheets of thickness `d` and `3d`, are touching each other. The temperature just outside the thinner sheet is `T_1` and on the side of the thicker sheet is `T_3`. The interface temperature is `T_2. T_1, T_2 and T_3` are in arithmetic progression. The ratio of thermal conductivity of thinner sheet to thicker sheet is

A

`1:3`

B

`3:1`

C

`2:3`

D

`3:9`

Text Solution

Verified by Experts

The correct Answer is:
A

`H_1 = H_2`
`:. (T_1-T_2)/((d//K_1A)) = (T_2-T_3)/((3d//K_2A))`
`K_1/K_2 = ((T_2-T_3)/(T_1-T_2))*1/3`
But `T_1: T_2 and T_3` are in AP
`:. T_2-T_3 = T_1 - T_2`
`:. K_1/K_2 = 1/3` .
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