Q. The dimensions of coefficient of thermal conductivity is
A) $ML^{2}T^{- 2}K^{- 1}$
B) $MLT^{- 3}K^{- 1}$
C) $MLT^{- 2}K^{- 1}$
D) $MLT^{- 3}K$
Explanation
$\frac{dQ}{dt} = - KA\left( \frac{d\theta}{dx} \right)$ $\Rightarrow \lbrack K\rbrack = \frac{\left\lbrack ML^{2}T^{- 2} \right\rbrack}{\lbrack T\rbrack} \times \frac{\lbrack L\rbrack}{\left\lbrack L^{2} \right\rbrack\lbrack K\rbrack} = MLT^{- 3}K^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.