Question

Watt per metre Kelvin is the unit of __________.
A) stress
B) period of revolution of a satellite
C) angular displacement
D) coefficient of thermal conductivity

Answer 1

Hint: Watt is the unit of power and power is defined as the work or energy over time. Metre is the unit of length and Kelvin is the unit of temperature. We can employ dimensional analysis to determine which of the given physical quantities has the unit of Watt per metre Kelvin. Among the given physical quantities, only the coefficient of thermal conductivity is explicitly dependent on the temperature as it is the property of a material by which it transfers heat from one point to another.

Formula used:
-The coefficient of thermal conductivity is given by, $\kappa ={\displaystyle \frac{HL}{A\mathrm{\Delta }T}}$ where $H$ is the heat current, $L$ is the length of the body, $A$ is the area of cross-section of the body and $\mathrm{\Delta }T$ is the temperature difference between two points in the body.
-The power is given by, $P={\displaystyle \frac{Fd}{t}}$ where $F$ is the force acting on the body, $d$ is its displacement and $t$ is the time.

Complete step by step solution:
Step 1: Express the dimensions of the physical quantity having the given unit.
The given unit is Watt per metre Kelvin i.e., $\text{W}{\text{m}}^{-1}{\text{K}}^{-1}$ or ${\displaystyle \frac{\text{W}}{\text{mK}}}$ ------- (1).
We know that power has the unit Watt, and power can be expressed as $P={\displaystyle \frac{W}{t}}={\displaystyle \frac{Fd}{t}}$ where $F$ is the force, $d$ is the displacement and $t$ is the time.
Then in dimensions, the unit Watt can be expressed as $\text{W}\to {\displaystyle \frac{ML{T}^{-2}}{T}}=ML{T}^{-3}$ ------ (2)
The unit of metre has the dimension of length and so we have $\text{m}\to L$ ------ (3)
Now expressing equation (1) in terms of the dimensions given by equations (2) and (3) we get, ${\displaystyle \frac{\text{W}}{\text{mK}}}\to {\displaystyle \frac{ML{T}^{-3}}{LK}}=M{T}^{-3}{K}^{-1}$
Thus the dimensions of the unknown quantity are $M{T}^{-3}{K}^{-1}$ .
Step 2: Express the dimensions of the coefficient of thermal conductivity.
The coefficient of thermal conductivity can be expressed as $\kappa ={\displaystyle \frac{HL}{A\mathrm{\Delta }T}}$ --------- (2) where $H$ is the heat current, $L$ is the length of the body, $A$ is the area of cross-section of the body and $\mathrm{\Delta }T$ is the temperature difference between two points in the body.
The dimension of heat current is $H\to ML{T}^{-3}$ .
The dimension of length is well $L$ .
The dimension of the area is $A\to L^2$ .
And the dimension of temperature difference is $\mathrm{\Delta }T\to K$ .
Now in terms of dimensions, the coefficient of thermal conductivity becomes $\kappa \to {\displaystyle \frac{ML{T}^{-3}L}{L^{2}K}}=M{T}^{-3}{K}^{-1}$
Thus the dimensions of the given unit and that of the coefficient of thermal conductivity match.

So the correct option is D.

Note: Alternate method
The coefficient of thermal conductivity can be expressed as $\kappa ={\displaystyle \frac{HL}{A\mathrm{\Delta }T}}$ --------- (A) where $H$ is the heat current, $L$ is the length of the body, $A$ is the area of cross-section of the body and $\mathrm{\Delta }T$ is the temperature difference between two points in the body.
The unit of heat current is Joule per second $\text{J}{\text{s}}^{-1}$ , the unit of length is metre $\text{m}$ , the unit of cross-sectional area is squared metre ${\text{m}}^2$ and the unit of temperature difference is Kelvin $\text{K}$ .
Now substituting the units of each quantity in equation (A) we get, $\kappa ={\displaystyle \frac{\text{J}{\text{s}}^{-1}\text{m}}{{\text{m}}^2\text{K}}}=\text{J}{\text{s}}^{-1}{\text{m}}^{-1}{\text{K}}^{-1}$
Now $\text{J}{\text{s}}^{-1}$ refers to Watt as Joule is the unit of energy or work, second is the unit of time and power which has the unit of Watt is defined as work or energy over time $P={\displaystyle \frac{W}{t}}$ .
Thus we have the unit of coefficient of thermal conductivity as $\text{W}{\text{m}}^{-1}{\text{K}}^{-1}$ .