Use Fourier’s law of heat conduction

Thermal conductivity $k$ is defined by $\frac{Q}{t} = k A \frac{\Delta T}{L}$ where:

• $\frac{Q}{t}$ = heat flow per unit time = power
• $A$ = area
• $\Delta T$ = temperature difference
• $L$ = length

Write dimensions of each quantity

• Heat (energy): $[Q] = [ML^2T^{-2}]$
• Therefore, $\left[\frac{Q}{t}\right] = [ML^2T^{-3}]$
• Area: $[A] = [L^2]$
• Length: $[L] = [L]$
• Temperature difference: $[\Delta T] = [K]$

Find dimensions of thermal conductivity $k$

From $\frac{Q}{t} = k A \frac{\Delta T}{L}$ we get $[k] = \frac{\left[Q/t\right]\,[L]}{[A][K]}$

Substitute dimensions: $[k] = \frac{[ML^2T^{-3}]\,[L]}{[L^2][K]}$

$[k] = [ML^{3-2}T^{-3}K^{-1}]$

$[k] = [MLT^{-3}K^{-1}]$

Match with options

The dimensional formula is $MLT^{-3}K^{-1}$

So the correct option is A.