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JEE Main 2020 (4 Sep, Shift 1) · Units & Dimensions

Units & Dimensions question from JEE Main 2020 (4 Sep, Shift 1)

Q71JEE Main 2020 · 4 SEP · SHIFT 1Easy

Units & Dimensions

Dimensional formula for thermal conductivity is (here K denotes the temperature):

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Answer: (a)

Heat flow: $\frac{Q}{t} = kA\frac{\Delta\theta}{x}$ , so $k = \frac{(Q/t)\,x}{A\,\Delta\theta}$ .
$[k] = \frac{\mathrm{ML^2T^{-3}}\cdot\mathrm{L}}{\mathrm{L^2}\cdot\mathrm{K}} = \mathrm{MLT^{-3}K^{-1}}$
So the answer is (a).

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