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JEE Main 2024 (31 Jan, Shift 2) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2024 (31 Jan, Shift 2)

Q22JEE Main 2024 · 31 JAN · SHIFT 2Easy

Applications of Dimensional Analysis

Consider two physical quantities $A$ and $B$ related to each other as $E = \frac{B - x^2}{At}$ where $E, x$ and $t$ have dimensions of energy, length and time respectively. The dimension of $AB$ is

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Answer: (b)

$B$ is subtracted from $x^2$ : $[B] = [L^2]$ .
$[E] = [L^2]/([A][T]) \Rightarrow [A] = [L^2]/([ML^2T^{-2}][T]) = [M^{-1}T]$ .
$[AB] = [L^2M^{-1}T^1]$ .
So the answer is (b).

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