JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 2)
Dimensional formula for thermal conductivity is (here K denotes the temperature):
MLT–3K–1
MLT–2K–2
MLT–2K
MLT–3K
Explanation
$$ \therefore $$ $${{d\theta } \over {dt}} = kA{{dT} \over {dx}}$$
$$ \Rightarrow $$ k = $${{\left( {{{d\theta } \over {dt}}} \right)} \over {A\left( {{{dT} \over {dx}}} \right)}}$$
$$ \Rightarrow $$ [k] = $${{\left[ {M{L^2}{T^{ - 3}}} \right]} \over {\left[ {{L^2}} \right]\left[ {K{L^{ - 1}}} \right]}}$$
= [MLT–3K–1]
$$ \Rightarrow $$ k = $${{\left( {{{d\theta } \over {dt}}} \right)} \over {A\left( {{{dT} \over {dx}}} \right)}}$$
$$ \Rightarrow $$ [k] = $${{\left[ {M{L^2}{T^{ - 3}}} \right]} \over {\left[ {{L^2}} \right]\left[ {K{L^{ - 1}}} \right]}}$$
= [MLT–3K–1]
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