The ends of the two rods of conductivities , radii and lengths in the ratio of `1:2` are maintained at the same temperature difference. If the rate of flow through the bigger rod is `12" cal s"^(-1)`, in shorter it will be (in `"cal s"^(-1)`) :

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

`H=(Q)/(t)=(kA(T_(1)-T_(2)))/(d)`
`therefore H_(1)=(k_(1)A_(1)(T_(1)-T_(2)))/(d_(1))`
`H_(2)=(k_(2)A_(2)(T_(1)-T_(2)))/(d_(2))`
`:.(H_(1))/(H_(2))=(k_(1))/(k_(2))*(A_(1))/(A_(2))xx(d_(2))/(d_(1))=(1)/(2)xx(1/2)^(2)xx2/1=(1)/(4)`
`:.H_(1)=H_(2)xx1/4=12xx1/4=3" cal s"^(-1)`.
Thus correct choice is (c ).