If speed (V), acceleration (A) and force (F) | vedclass

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Question

$\begin{array}{l}
\frac{F}{A} = y \cdot \frac{{\Delta \ell }}{\ell }\,\,;\,\,\left[ Y ight] = \frac{F}{A}\
Now\,from\,\dim ension\
\,\,\,\,\,\,

Vedclass · Accepted Answer

${V^{ - 4}}{A^{2}}F$

Vedclass · Answer

$\begin{array}{l}
\frac{F}{A} = y \cdot \frac{{\Delta \ell }}{\ell }\,\,;\,\,\left[ Y ight] = \frac{F}{A}\
Now\,from\,\dim ension\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F = \frac{{ML}}{{{T^2}}}\,\,;\,\,L = \frac{F}{M} \cdot {T^2}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{L^2} = \frac{{{F^2}}}{{{M^2}}}{\left( {\frac{V}{A}} ight)^4}\,\,\,\,\,\,T = \frac{V}{A}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{L^2} = \frac{{{F^2}}}{{{M^2}{A^2}}}\frac{{{V^4}}}{{{A^2}}}\,\,\,\,\,\,F = MA\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,{L^2} = \frac{{{V^4}}}{{{A^2}}}\
\left[ Y ight] = \frac{{\left[ F ight]}}{{\left[ A ight]}} = {F^1}{V^{ - 4}}{A^2}
\end{array}$

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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