A) \[M{{L}^{2}}{{T}^{-2}}{{K}^{-1}}\]
B) \[ML{{T}^{-3}}{{K}^{-1}}\]
C) \[ML{{T}^{-2}}{{K}^{-1}}\]
D) \[ML{{T}^{-3}}K\]
Correct Answer: B
Solution :
\[\frac{dQ}{dt}=-KA\left( \frac{d\theta }{dx} \right)\] \[\Rightarrow \] [K] = \[\frac{[M{{L}^{2}}{{T}^{-2}}]}{[T]}\,\times \,\frac{[L]}{[{{L}^{2}}]\ [K]}\]=\[ML{{T}^{-3}}{{K}^{-1}}\]
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