Solve this following
Question:

Dimensional formula for thermal conductivity is (here $\mathrm{K}$ denotes the temperature)

 

  1. $\mathrm{MLT}^{-3} \mathrm{~K}$

  2. $\mathrm{MLT}^{-2} \mathrm{~K}$

  3. $\mathrm{MLT}^{-2} \mathrm{~K}^{-2}$

  4. $\mathrm{MLT}^{-3} \mathrm{~K}^{-1}$


Correct Option: , 4

Solution:

$\because \frac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{kA} \frac{\mathrm{dT}}{\mathrm{dx}}$

$\mathrm{k}=\frac{\left(\frac{\mathrm{d} \theta}{\mathrm{dt}}\right)}{\mathrm{A}\left(\frac{\mathrm{dT}}{\mathrm{dx}}\right)}$

$[\mathrm{k}]=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]}{\left[\mathrm{L}^{2}\right]\left[\mathrm{KL}^{-1}\right]}=\left[\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\right]$

 

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