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)`

Answer

B

Explanation

Rate of heat flow –
`((Q)/(t))=(kA triangle T)/(l) implies Q/t prop (r^2)/(l)`
` triangle (Q_1)/(Q_2)=((r_1)/(r_2))^2((l_2)/(l_1))=(1/2)^2 xx 2/1 =1/2 implies Q_2 =2Q_1`.