A rod `CD` of thermal resistance `5.0KW_(-1)` is joined at the middle of an identical rod `AB` as shown in figure. The ends `A`,` B` and `D` are maintained at `100^(@)C` , `0^(@)C` and `25^(@)C` respectively. Find the heat current in `CD`.

Options

A. `35^(@)C, 4W`
B. `45^(@)C, 4W`
C. `35^(@)C, 3W`
D. `45^(@)C, 3W`

Answer

B

Explanation

(2) `R_(AC) = R_(1) = (l//2)/(KA) = (R )/(2)`
`R_(CB) = R_(2) = (l//2)/(KA) = (R )/(2)`
`R_(CD) = R_(3) = (l)/(KA) = R`
Let temperature of `C` is `theta`
`i = i_(1) + i_(2)`
`(100 - theta)/(R_(1)) = (theta - 25)/(R_(3)) + (theta - 0)/(R_(2))`
`(100 - theta)/(R//2) = (theta - 25)/(R ) + (theta)/(R // 2)`
`200 - 2 theta = theta - 25 + 2 theta`
`5 theta 225 implies theta = 45^(@)C`
Heat current in `CD`
`i_(1) = (theta - 25)/(R ) = (45 - 25)/(5) = 4 W`