Question:
Dimensional formula for thermal conductivity is (here $\mathrm{K}$ denotes the temperature)
Correct Option: , 4
Solution:
$\because \frac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{kA} \frac{\mathrm{dT}}{\mathrm{dx}}$
$\mathrm{k}=\frac{\left(\frac{\mathrm{d} \theta}{\mathrm{dt}}\right)}{\mathrm{A}\left(\frac{\mathrm{dT}}{\mathrm{dx}}\right)}$
$[\mathrm{k}]=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]}{\left[\mathrm{L}^{2}\right]\left[\mathrm{KL}^{-1}\right]}=\left[\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\right]$