Six identical cunducting rods are joined as shown in Fig. Points A and D are maintained at temperatures `200^@C` and `20^@C` respectively. The temperature of junction B will be

Options

A. `120^(@)C`
B. `100^(@)C`
C. `140^(@)C`
D. `80^(@)C`

Answer

C

Explanation

`H_(1) = 2H_(2)`
`(KA)/(l) (200-T)=2(KA)/(2l)(T-T')`
`200=2T-T'"...."(1)`
`2H_(2) = H_(3)`
`2(KA)/(2l) (T-T') = (T'-20) (KA)/(l)`
`2T' -T=20`
By solving `(1)` & (2)
`T = 140`