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Question: The dimensions of coefficient of thermal conductivity is...

The dimensions of coefficient of thermal conductivity is

A

ML2T−2K−1ML^{2}T^{- 2}K^{- 1}

B

MLT−3K−1MLT^{- 3}K^{- 1}

C

MLT−2K−1MLT^{- 2}K^{- 1}

D

MLT−3KMLT^{- 3}K

Answer

MLT−3K−1MLT^{- 3}K^{- 1}

Explanation

Solution

dQdt=−KA(dθdx)\frac{dQ}{dt} = - KA\left( \frac{d\theta}{dx} \right)

⇒\Rightarrow [K] = [ML2T−2][T]×[L][L2] 6mu[K]\frac{\lbrack ML^{2}T^{- 2}\rbrack}{\lbrack T\rbrack} \times \frac{\lbrack L\rbrack}{\lbrack L^{2}\rbrack\mspace{6mu}\lbrack K\rbrack}=MLT−3K−1MLT^{- 3}K^{- 1}