InterviewSolution
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InterviewSolution
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A uniform metal ring with centre `C` have two points `A` and `B` such that angle `ACB` is `theta`. `A` and `B` are maintained at two different constant temperature. If the angle between `A` and `B` i.e. `theta=180^(@)` the rate of heat flow from `A` and`B` is `1.2W`, then what will be the rate, when `theta=90^(@)`?A. `0.6W`B. `0.9W`C. `1.6W`D. `1.8W` |
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Answer» (c) Let `DeltaT=` temperature difference between the rings `R=` total internal resistance of the ring when `theta=180^(@)` The resistance will be `R//2, R//2` and both are in parallel so, equivalent resistance `=R//4` Rate of total heat flow `=(Q_(1))/t=1.2=(DeltaT)/((R//4))implies` When `theta=90^(@)` there are two sections with resistances `R//4` adn `3R//4` in parallel. So, equivalent resistance `=(3R)/16` Rate of heat flow `=(Q_(2))/t=(DeltaT)/(((3R)/16))` `l_(2)=16/3((DeltaT)/R)` `l_(2)=1.6W` `impliesl_(2)=1.6W` |
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