Question

The dimension of coefficient of thermal conductivity is
$(\frac{dQ}{dt}=-KA\frac{dT}{dx})$, where $dQ$ is heat transferred, $dt$ is time taken, $K$ is coefficient of thermal conductivity, $A$ is cross-sectional area of conductor, $dT$ is change in temperature and $dx$ is length of conductor.

• A. $\left[M{L}^2{T}^{-2}{K}^{-1}\right]$
• B. $\left[ML{T}^{-3}{K}^{-1}\right]$
• C. $\left[ML{T}^{-2}{K}^{-1}\right]$
• D. $\left[ML{T}^{-3}K\right]$

Answer 1

The correct option is B $\left[ML{T}^{-3}{K}^{-1}\right]$

We know that the rate of heat transfer is given by,
$H=\frac{dQ}{dt}=\frac{-KAdT}{dx}$
$\Rightarrow K=\frac{dQdx}{AdTdt}$
$\Rightarrow \left[K\right]=\left[\frac{dQdx}{AdTdt}\right]$
$\Rightarrow \left[K\right]=\left[\frac{\left(M{L}^2{T}^{-2}\right)\times \left(M^{0}L{T}^0\right)}{\left(M^0{L}^2{T}^0\right)\times \left(M^0{L}^0{T}^{0}K\right)\times \left(M^0{L}^0{T}^1\right)}\right]$
$\Rightarrow \left[K\right]=\left[ML{T}^{-3}{K}^{-1}\right]$