i. Thermal resistance is defined as `R=l//kA`
For rod A `R_A=(l)/(k_(A)A)=(l)/(3kA)`
For rod B `R_B=(l)/(k_(B)A)=(l)/(kA)`
ii. Using electrical analogy, the given problem may be modified as The equivalent resistance is `R=R_A+R_B=(4)/(3)(l)/(kA)`
The heat current is given by
`I=(T_1-T_2)/(R )=(100-0)/((4)/(3)(l)/(kA))=73((kA)/(l))`
iii. The temperature of the junction is given by
`T=T_1-IR_A=100-(75kA)/(l)((l)/(3kA))=75^@C`
iv. The variation of temperature along the length of the conductor is shown in Fig.
