Question

Three rods each of same length and cross - section are joined in series. The thermal conductivity of the materials are K, 2K and 3K respectively. If one end is kept at `200^@C` and the other at `100^@C`. What would be the temperature of the junctions in the steady state? Assume that no heat is lost due to radiation from the sides of the rods.

Answer 1

Correct Answer - A::B::C::D

`H_(1)=H_(2)=H_(3)`
`:. (200-theta_(1))/((1//KA))=(theta_(1)-theta_(2))/((l//2KA))=(theta_(2)-100)/((l//3KA))`
Solving these equations we get,
`theta_1 = 145.5^@C`
and `theta_2 = 118.2^@C` .