# If speed $\mathrm{V}$, area $\mathrm{A}$ and force $\mathrm{F}$ are chosen as fundamental units, then the dimension of Young’s modulus will be :

Question:

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

1. $\mathrm{FA}^{2} \mathrm{~V}^{-1}$

2. $\mathrm{FA}^{2} \mathrm{~V}^{-3}$

3. $\mathrm{FA}^{2} \mathrm{~V}^{-2}$

4. $\mathrm{FA}^{-1} \mathrm{~V}^{0}$

JEE Main Previous Year Single Correct Question of JEE Main from Chemistry Laws of Motion chapter.

JEE Main Previous Year 2020

Correct Option: 4

Solution:

Young’s modulus, $Y=\frac{\text { stress }}{\text { strain }}$

$\Rightarrow Y=\frac{\mathrm{F}}{\mathrm{A}} / \frac{\Delta \ell}{\ell_{0}}=\mathrm{FA}^{-1} \mathrm{~V}^{0}$

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