JEE PYQ: Units And Dimensions Question 9
Question 9 - 2020 (04 Sep Shift 1)
Dimensional formula for thermal conductivity is (here $K$ denotes the temperature):
(1) $MLT^{-2}K$
(2) $MLT^{-2}K^{-2}$
(3) $MLT^{-3}K$
(4) $MLT^{-3}K^{-1}$
Show Answer
Answer: (4)
Solution
From formula, $\frac{dQ}{dt} = kA\frac{dT}{dx}$. So $k = \frac{[ML^2T^{-3}]}{[L^2][KL^{-1}]} = [MLT^{-3}K^{-1}]$.