A cylindrical rod is having temperature `T_(1)` and `T_(2)` at its ends. The rate of flow of heat is `Q_(1)` cal/sec. If all the linear dimensions are doubled keeping temperatures constant, then 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)="k A"((T_(1)-T_(2))t)/(d)`
When linear diamension are doubled then A becomes 4 times and d becomes 2 times.
`:.(dQ)/(dt)` becomes 2 times
Correct choice is (c ).