(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`