Q. Two cylindrical conductors $A\ and\ B$ of same metallic material have their diameters in the ratio $1\ :2$ and lengths in the ratio$2\ :1$. If the temperature difference between their ends is same, the ratio of heat conducted respectively by $A\ and\ B\ $per second is
A) 1 : 2
B) 1 : 4
C) 1 : 16
D) 1 : 8
Explanation
(d) Here, $\frac{D_{1}}{D_{2}} = \frac{1}{2}\ $
$\frac{A_{1}}{A_{2}} = \frac{D_{1}^{2}}{D_{2}^{2}} = \frac{1}{4}$
$\frac{dx_{1}}{dx_{2}} = \frac{2}{1}$
$\frac{dQ_{1}}{dt} = KA_{1}\frac{dT}{dx_{1}}\ :\frac{dQ_{2}}{dt} = KA_{2}\frac{dT}{dx_{2}}$
$\frac{dQ_{1}/dt}{dQ_{2}/dt} = \frac{A_{1}}{dx_{1}}.\frac{dx_{2}}{A_{2}} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.