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JEE Main 2020 (2 Sep, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2020 (2 Sep, Shift 1)

Q54JEE Main 2020 · 2 SEP · SHIFT 1Easy

Applications of Dimensional Analysis

If speed V, area A and force F are chosen as fundamental units, then the dimension of Young’s modulus will be

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Answer: (a)

Young's modulus $Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{F}{A}$ (strain is dimensionless).
So $[Y] = FA^{-1}V^{0}$ .
So the answer is (a).

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