A cylindrical rod 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 temperature constant, then rate of flow of heat `Q_(2)` will be

Options

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

Explanation

(b) Initially, `Q_(1)=(KA_(1)(T_(1)-T_(2)))/(l_(1))` but on doubling all
dimensions `l_(2)=2l_(1)` and `A_()=4A_(1)`
Hence `Q_(2)=(KA_(2)(T_(1)-T_(2)))/(l_(2))=(K4A_(1)(T_(1)-T_(2)))/(2l_(1))`
`=2(KA_(1)(T_(1)-T_(2)))/(l_(1))=2Q_(1)`