Question

Rate of heat flow through a cylindrical rod is `H_(1)`. Temperatures of ends of rod are `T_(1)` and `T_(2)`. If all the dimensions of rod become double and temperature difference remains same and rate of heat flow becomes `H_(2)`. Then `(H_(1))/(H_(2))` is `0.x`. Find value of x.
A. `H_(2) = 2H_(1)`
B. `H_(2) = H_(1)`
C. `H_(2) = H_(1)/4`
D. `H_(2) = 4H_(1)`

Answer 1

Correct Answer - A
Rate of heat flow, `H = Q/t = (KAT_(1)-T_(2))/l`
or `H alphaA/l` or `H_(2)=2H_(1)`