Q. The ends of 2 different materials with their thermal conductivities, radii of cross section and length all in the ratio of $1\ :2$ maintained at temperature difference. If the rate of the flow of heat in the longer rod is 4 cal$s^{- 1}$, that in the shorter rod in cal$s^{- 1}$ will be
A) 1
B) 2
C) 8
D) 6
Explanation
(a) Here, $\frac{K_{1}}{K_{2}} = \frac{1}{2},\frac{r_{1}}{r_{2}} = \frac{1}{2}\ $
$\therefore\ \frac{A_{1}}{A_{2}} = \frac{1}{4}$
$\frac{dx_{1}}{dx_{2}} = \frac{1}{2},\frac{dQ_{2}}{dt} = 4\ cals^{- 1}\ ,\ \ \frac{dQ_{1}}{dt} = \ ?$
$\frac{dQ_{2}/dt}{dQ_{1}/dt} = \ \frac{K_{2}A_{2}dT/dx_{2}}{K_{1}A_{1}dT/dx_{1}} = \frac{K_{2}}{K_{1}}\frac{A_{2}}{A_{1}}\frac{dx_{1}}{dx_{2}}$
= $2 \times 4 \times \frac{1}{2} = 4$
$\frac{dQ_{1}}{dt} = \ \frac{dQ_{2}/dt}{4} = \frac{4}{4} = 1\ cals^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.