Question

A wire consists of two layers as shown in the figure. Thermal conductivity of inner core of radius $2\text{cm}$ is $k$ and of the outer one of radius $4\text{cm}$ is $2k$. Find the equivalent thermal conductivity between its two ends $a$ and $b\left(k_{\text{eq}}\right)$.

• A. $\frac{7k}{2}$
• B. $\frac{7k}{8}$
• C. $\frac{7k}{4}$
• D. $\frac{5k}{4}$

Answer 1

The correct option is C $\frac{7k}{4}$

Given,
Radius of inner core $\left(r_1\right)=2\text{cm}$
Radius of outer core $\left(r_2\right)=4\text{cm}$
Thermal conductivity of inner core $\left(k_1\right)=k$
Thermal conductivity of outer core $\left(k_2\right)=2k$
Between two ends $a$ and $b$, two cylinders are in parallel to each other.
$\therefore \frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}$ (where $R_1,R_2$ are thermal resistance of inner and outer core)
Since, $R=\frac{l}{kA}$ we can deduce that
$\frac{k_{\text{eq}}A}{l}=\frac{k_1{A}_1}{l_1}+\frac{k_2{A}_2}{l_2}$
$\frac{k_{\text{eq}}\left(\pi \right(r_2{)}^2)}{l}=\frac{k_1\left(\pi r_1^2\right)}{l}+\frac{k_2(\pi r_2^2-\pi r_1^2)}{l}$
$16k_{\text{eq}}=4k+2k\left(12\right)$
$k_{\text{eq}}=\frac{28}{16}k\Rightarrow k_{\text{eq}}=\frac{7}{4}k$