Question

What is dimensional formula of thermal conductivity ?

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

Answer 1

Answer: (B)

Substituting dimensions for corresponding quantities in the relation having coefficient of thermal conductivity.
Heat $\Delta Q$ transferred through a rod of length $L$ and area $A$ in time $\Delta t$
$\Delta Q=KA\left(\frac{T_1-T_2}{L}\right)\Delta t$
where $K=$ coefficient of thermal conductivity, $T_1-T_2=$ temperature different .
$\therefore k=\frac{\Delta Q\times L}{A\left(T_1-T_2\right)\Delta t}...\left(i\right)$
Substituting dimensions for corresponding quantities in Eq. $\left(i\right)$, we have
$\left[K\right]=\frac{\left[M{L}^2{T}^{-2}\right]\left[L\right]}{\left[L^2\right]\left[\theta \right]\left[T\right]}$
$=\left[ML{T}^{-3}{\theta }^{-1}\right]$