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

Options

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

Answer

C

Explanation

`H_(1)=KA_(1)((T_(1)-T_(2))/(L_(1)))`
`H_(2)=KA_(2)((T_(1)-T_(2))/(L_(2)))`
`:.(H_(1))/(H_(2))=(A_(1))/(A_(2))((L_(2))/(L_(1)))`
`=((pid_(1)^(2))/(4))/((pid_(2)^(2))/(4))*(L_(2))/(L_(1))`
`=(d_(1)^(2))/(d_(2)^(2))xx(L_(2))/(L_(1))`
`:.(Q_(1))/(Q_(2))=(d_(1)^(2))/((2d_(1))^(2))xx(2L_(1))/(L_(1))`
`=1/4xx2`
`=1/2`
`:.Q_(2)=2Q_(1)`