Q 11-01-209JEE MainJEE Main 2020 (4 Sep, Shift 1)Easy
Dimensional formula for thermal conductivity is (here $K$ denotes the temperature):
Answer: (D) $MLT^{-3}K^{-1}$
Heat current $\dfrac{dQ}{dt} = kA\dfrac{\Delta T}{l}$, so
$$k = \frac{(dQ/dt)\,l}{A\,\Delta T} = \frac{ML^{2}T^{-3}\cdot L}{L^{2}\cdot K} = MLT^{-3}K^{-1}$$
Solution by Sreeraj P, M.Sc Physics