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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 :A. `0.6W`B. `0.9W`C. `1.6 W`D. `1.8W` |
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Answer» Correct Answer - C 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` . |
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