Question

A composite rod made of three rods of equal length and cross-section as shown in the figure. The thermal conductivities of the materials of the rods are $K/2,5K$ and K respectively. The end A and end B are at constant temperatures. All heat entering the face A goes out of the end B. Assuming no loss of heat from the sides of the bar. The effective thermal conductivity of the bar is:

A.)$\dfrac{{15K}}{{16}}$
B.)$\dfrac{{6K}}{{13}}$
C.)$\dfrac{{5K}}{{16}}$
D.)$\dfrac{{2K}}{{13}}$

Answer 1

For solving this numerical we will understand the thermal conductivity first and then the equivalent thermal conductivity of three rods by the given formula.

${\text{K = }}\dfrac{{{\text{(QL)}}}}{{\left( {{\text{A}}\Delta {\text{T}}} \right)}}$

Where, $K$ is the thermal conductivity, $Q$ is the amount of heat transmitted in Joules per second or Watts through the material.

Complete step-by-step answer:

Let ${k_1}$ and ${k_2}$ be thermal conductivities of two resistors. For resistors connected in series, equivalent thermal conductivity is,

${k_s}{\text{ = }}\dfrac{{2{k_1}{k_2}}}{{\left( {{k_1} + {k_2}} \right)}}{\text{ }}$

For resistors connected in parallel, equivalent thermal conductivity is $kp = {k_1} + {k_2}$.

Similarly for three rods,

$\dfrac{{3l}}{K} = \dfrac{l}{{k/2}} + \dfrac{l}{{5k}} + \dfrac{l}{k}$

$\Rightarrow \dfrac{{3l}}{K} = \dfrac{{2l}}{k} + \dfrac{l}{{5k}} + \dfrac{l}{k}$

$\Rightarrow \dfrac{{3l}}{K} = \dfrac{{16l}}{{5k}}$

$\therefore K = \dfrac{{15k}}{{16}}$

Hence, the thermal conductivities of the three resistors in series is $K = \dfrac{{15k}}{{16}}$.

Hence, the correct option is A.

Additional Information: Thermal conductivity, as a temperature gradient exists perpendicular to the surface, can be defined as the rate at which heat is transmitted by conduction through a unit cross-sectional region of a substance. A material's thermal conductivity is defined by the following formula:

${\text{K = }}\dfrac{{{\text{(QL)}}}}{{\left( {{\text{A}}\Delta {\text{T}}} \right)}}$

where $K$ is the thermal conductivity. $Q$ is the amount of heat transmitted in Joules per second or Watts through the material.

Note: Thermal conductivity is a material property. It will not differ with the dimensions of a material, but it is dependent on the temperature, the density and the moisture content of the material. The thermal conductivity of a material depends on its temperature, density and moisture content.