A cylindrical rod is having temperature `T_(1)` and `T_(2)` at its ends. The rate of flow of heat is `Q_(1)` cal/s. If all the linear dimensions are doubled keeping temperatures constant, the rate of flow of heat `Q_(2)` will be :

Options

A. `4Q_(1)`
B. `Q_(1)//4`
C. `2Q_(1)`
D. `Q_(1)//2`.

Answer

C

Explanation

`(dQ)/(dt)=(KA(T_(1)-T_(2)))/(l)=(Kpir^(2)(T_(1)-T_(2)))/(l)=Q_(1)`
When all dimensions are doubled then `(dQ)/(dt)` becomes two times i.e. `2Q_(1)`
Correct choice is (c ).