Question

Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at `0^@C` and `90^@C`, respectively. The temperature of junction of the three rods will be
(a) `45^@C` (b) `60^@C`
(c) `30^@C` (d) ` 20^@C`.

A. `45^(@)C`
B. `60^(@)C`
C. `30^(@)C`
D. `20^(@)C`

Answer 1

`(b)` Let `theta^(@)C` be the temperature at `B`. Let `Q` is the heat flowing per second from `A` to `B` on account of temperature difference by conductivity.
`:. Q=(KA(90-theta))/(l)`…….`(i)`
Where `K=` Thermal conducitivity of the rod
`A=` Area of cross-section of the rod
`l=` Length of the rod
By symmetry, the same will be the case for heat flow from `C` to `B`
`:.` The heat flowing per second from `B` to `D` will be
`2Q=(KA(theta-0))/(l)`......`(ii)`
Diveding equation `(ii)` by equation `(i)`
`2=(theta)/(92-theta)`
`implies theta=60^(@)`