Two plates of thermal conductivities $k_1$ and $k_2$,cross-sectional areas $A_1$ and $A_2$,and the same thickness $l$ are joined as shown in the figure. The equivalent thermal conductivity $k$ of the combination is:

  • A
    $k_1 A_1 + k_2 A_2$
  • B
    $\frac {k_1 A_1}{ k_2 A_2}$
  • C
    $\frac{k_1 A_1 + k_2 A_2}{A_1 + A_2}$
  • D
    $\frac{k_1 A_2 + k_2 A_1 }{k_1 + k_2}$

Explore More

Similar Questions

An iron bar $(L_{1}=0.1 \; m, A_{1}=0.02 \; m^{2}, K_{1}=79 \; W m^{-1} K^{-1})$ and a brass bar $(L_{2}=0.1 \; m, A_{2}=0.02 \; m^{2}, K_{2}=109 \; W m^{-1} K^{-1})$ are soldered end to end as shown in the figure. The free ends of the iron bar and brass bar are maintained at $373 \; K$ and $273 \; K$ respectively. Obtain expressions for and hence compute:
$(i)$ the temperature of the junction of the two bars,
$(ii)$ the equivalent thermal conductivity of the compound bar,and
$(iii)$ the heat current through the compound bar.

Five wires each of cross-sectional area $A$ and length $l$ are combined as shown. The thermal conductivity of copper and steel are $k_1$ and $k_2$ respectively. The equivalent thermal resistance between $A$ and $C$ is

Difficult
View Solution

Four rods of the same material and having the same cross-section and length have been joined as shown in the figure. The temperature of the junction of the four rods will be ............... $^{\circ}C$.

$A$ rod $CD$ of thermal resistance $10.0 \; K/W$ is joined at the middle of an identical rod $AB$ as shown in the figure. The ends $A$,$B$,and $D$ are maintained at $200^{\circ}C$,$100^{\circ}C$,and $125^{\circ}C$ respectively. The heat current in $CD$ is $P$ watt. The value of $P$ is ... .

In a composite slab, there are two materials having coefficients of thermal conductivity $K$ and $2K$, and thicknesses $x$ and $4x$ respectively. The temperatures of the two outer surfaces of the composite slab are $T_2$ and $T_1$ $(T_2 > T_1)$. $T_2$ is on the side with conductivity $K$ and $T_1$ is on the side with conductivity $2K$. The rate of heat transfer through the slab in a steady state is $\left[ \frac{A(T_2 - T_1)K}{x} \right] \cdot f$, where $f$ is equal to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo