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1.Units, Dimensions and Measurement
medium
Dimensional formula for thermal conductivity is (here $K$ denotes the temperature)
A$MLT ^{-3} K$
B$MLT ^{-2} K$
C$ML T ^{-2} K ^{-2}$
D$MLT ^{-3} K ^{-1}$
Solution
$\because \frac{ d Q}{ dt }= k A \frac{ d T }{ d x }$
$k =\frac{\left(\frac{ d Q}{ dt }\right)}{ A \left(\frac{ d T }{ d x }\right)}$
$[ k ]=\frac{\left[ ML ^{2} T ^{-3}\right]}{\left[ L ^{2}\right]\left[ KL ^{-1}\right]}=\left[ MLT ^{-3} K ^{-1}\right]$
$k =\frac{\left(\frac{ d Q}{ dt }\right)}{ A \left(\frac{ d T }{ d x }\right)}$
$[ k ]=\frac{\left[ ML ^{2} T ^{-3}\right]}{\left[ L ^{2}\right]\left[ KL ^{-1}\right]}=\left[ MLT ^{-3} K ^{-1}\right]$
Standard 11
Physics
