If speed V,area A,and force F are chosen as f | vedclass

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(A) Let the dimension of Young's modulus $Y$ be expressed as $Y = F^{x} A^{y} V^{z}$.The dimensional formula for Young's modulus is $[M^{1} L^{-1} T^{

Vedclass · Accepted Answer

$FA^{-1}V^{0}$

Vedclass · Answer

(A) Let the dimension of Young's modulus $Y$ be expressed as $Y = F^{x} A^{y} V^{z}$.The dimensional formula for Young's modulus is $[M^{1} L^{-1} T^{-2}]$.The dimensional formula for Force $F$ is $[M^{1} L^{1} T^{-2}]$.The dimensional formula for Area $A$ is $[L^{2}]$.The dimensional formula for Speed $V$ is $[L^{1} T^{-1}]$.Substituting these into the equation:$[M^{1} L^{-1} T^{-2}] = [M^{1} L^{1} T^{-2}]^{x} [L^{2}]^{y} [L^{1} T^{-1}]^{z}$$[M^{1} L^{-1} T^{-2}] = [M]^{x} [L]^{x + 2y + z} [T]^{-2x - z}$Comparing the powers of $M$,$L$,and $T$ on both sides:For $M$: $x = 1$For $T$: $-2x - z = -2 \Rightarrow -2(1) - z = -2 \Rightarrow z = 0$For $L$: $x + 2y + z = -1 \Rightarrow 1 + 2y + 0 = -1 \Rightarrow 2y = -2 \Rightarrow y = -1$Thus,the dimension of Young's modulus is $F^{1} A^{-1} V^{0}$.

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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