Question

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 ${100}^{\circ }$ C​ 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. $\frac{1}{3}$

Answer 1

The correct option is A

$\frac{16}{3}$

Heat current $H=\frac{\Delta \theta }{R}\Rightarrow \frac{H_p}{H_s}=\frac{R_s}{R_p}$
In first case : $R_s=R_1+R_2=\frac{l}{\left(3K\right)A}+\frac{l}{KA}=\frac{4}{3}\frac{l}{KA}$
In second case : $R_p=\frac{R_1{R}_2}{R_1+R_2}=\frac{\frac{1}{\left(3K\right)A}\times \frac{l}{KA}}{(\frac{l}{\left(3K\right)A}+\frac{l}{KA})}=\frac{l}{4KA}$
$\therefore \frac{H_p}{H_s}=\frac{\frac{4l}{3KA}}{\frac{l}{4KA}}=\frac{16}{3}$