Question

The ends of a copper rod of length 1m and area of cross-section `1cm^2` are maintained at `0^@C` and `100^@C`. At the centre of the rod there is a source of heat of power 25 W. Calculate the temperature gradient in the two halves of the rod in steady state. Thermal conductivity of copper is `400 Wm^-1 K^-1`.

Answer 1

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

`H_1 + H_2 = 25`
`:. (theta -100)/(0.5//(400 xx 10^-4)) + (theta - 0)/(0.5 //(400 xx 10^-4)) = 25`
Solving this equation, we get
`theta ~~ 206^@C`
Now, temperature gradient on
`LHS = ((theta-100)/0.5) = 212^@C//m`
and on `RHS = (theta -0)/0.5 = 424 ^@C//m`.