Question:
If speed $(V)$, acceleration $(A)$ and force $(F)$ are considered as fundamental units, the dimension of Young's modulus will be :
If speed $(V)$, acceleration $(A)$ and force $(F)$ are considered as fundamental units, the dimension of Young's modulus will be :
Updated On: Jun 29, 2024
- $V^{-2} A^2 F^2$
- $V^{-4} A^2 F$
- $V^{-4} A^{-2} F$
- $V^{-2} A^2 F^{-2}$
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The Correct Option is B
Solution and Explanation
$\frac{F}{A} = y . \frac{\Delta\ell}{\ell} $
$ \left[Y\right] = \frac{F}{A}$
$ F = \frac{ML}{T^{2}} $
$ L = \frac{F}{M} .T^{2} $
$ L^{2} = \frac{F^{2}}{M^{2}} \left(\frac{V}{A}\right)^{4} \because T = \frac{V}{A}$
$ L^{2} = \frac{F^{2}}{M^{2}A^{2}} \frac{v^{4}}{A^{2}} F = MA $
$ L^{2} = \frac{V^{2}}{A^{2}} $
$ \left[Y\right] = \frac{\left[F\right]}{\left[A\right]} = F^{1} V^{-4} A^{2} $
$ \left[Y\right] = \frac{F}{A}$
$ F = \frac{ML}{T^{2}} $
$ L = \frac{F}{M} .T^{2} $
$ L^{2} = \frac{F^{2}}{M^{2}} \left(\frac{V}{A}\right)^{4} \because T = \frac{V}{A}$
$ L^{2} = \frac{F^{2}}{M^{2}A^{2}} \frac{v^{4}}{A^{2}} F = MA $
$ L^{2} = \frac{V^{2}}{A^{2}} $
$ \left[Y\right] = \frac{\left[F\right]}{\left[A\right]} = F^{1} V^{-4} A^{2} $
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