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`.

Options

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

Answer

B

Explanation

Let `theta^(@)C` be the temperature at `B` . Let `Q` be 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 conductivity of the rod, `A=` Area of cross section of the rod, `l=` length of the rod. By symmertry, the same will be the case for heat flow from
`C` to `B` .
`:.` The heat flowing per second from `B` top `D` will be
`2Q=(KA(theta-0))/(l)` (ii)
Dividing Eq. (ii) buy Eq. (i)
`2=(theta)/(90-theta) implies theta=60^(@)C`.