Q. Two conducting rods $A$ and $B$ of same length and cross-sectional area are connected (i) In series (ii) In parallel as shown. In both combination a temperature difference of $100℃$ is maintained. If thermal conductivity of $A$ is $3K$ and that of $B$ is $K$ then the ratio of heat current flowing in parallel combination to that flowing in series combination is
A) $\frac{16}{3}$
B) $\frac{3}{16}$
C) $\frac{1}{1}$
D) $\frac{1}{3}$
Explanation
(a) Heat current $H = \frac{\mathrm{\Delta}\theta}{R} \Rightarrow \frac{H_{P}}{H_{S}} = \frac{R_{S}}{R_{P}}$
In first case : $R_{S} = R_{1} + R_{2} = \frac{l}{(3K)A} + \frac{l}{KA} = \frac{4}{3}\frac{l}{KA}$
In second case : $R_{P} = \frac{R_{1}R_{2}}{R_{1} + R_{2}} = \frac{\frac{1}{(3K)A} \times \frac{l}{KA}}{\left( \frac{l}{(3K)A} + \frac{l}{KA} \right)} = \frac{l}{4KA}$
$\therefore\frac{H_{P}}{H_{S}} = \frac{\frac{4l}{3KA}}{\frac{l}{4KA}} = \frac{16}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.