The two ends of a metal rod are maintained at temperatures `100^(@)C` and `110^(@)C` . The rate of heat flow in the rod is found to be `4.0 j//s`. If the ends are maintaind at temperatures `200^(@)C` and `210^(@)C`, the rate of heat flow will be :

Options

A. `0.6W`
B. `0.9W`
C. `1.6 W`
D. `1.8W`

Answer

C

Explanation

R =Total thermal resistance of the ring `DeltaT` = difference in temperature between `A` and `B` for `theta =180^(@)` equivalent resistance between `A` and `B` is `R//2 & R//2` in parallel
Rate of flow of heat `I_(1) = 1.2 = (DeltaT)/(R//4) :. (DeltaT)/(R) =0.3`
For `theta =90^(@)` equivalent resistance between `A` & `B` is `3R//16` (`R //4` and `3R//4` in parallel)
Rate of flow of heat `I_(2) = (DeltaT)/(3R//16) = (16)/(3) xx 0.3 =1.6W` .