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JEE Main 2019 (11 Jan, Shift 2) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2019 (11 Jan, Shift 2)

Q58JEE Main 2019 · 11 JAN · SHIFT 2Easy

Applications of Dimensional Analysis

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be:

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Answer: (c)

Young's modulus $Y$ = force/area: $Y = F/L^2$ .
Length in terms of speed and acceleration: $L = V^2/A$ , so $L^2 = V^4A^{-2}$ .
$Y = F V^{-4} A^2$ .
So the answer is (c) $V^{-4}A^2F$ .

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