If force (F),length (L),and time (T) are assu | vedclass

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(A) Let the mass $M$ be expressed as $M = k F^a L^b T^c$,where $k$ is a dimensionless constant.Substituting the dimensions of each quantity:$[M] = [M^

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$F L^{-1} T^2$

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(A) Let the mass $M$ be expressed as $M = k F^a L^b T^c$,where $k$ is a dimensionless constant.Substituting the dimensions of each quantity:$[M] = [M^1 L^0 T^0]$$[F] = [M^1 L^1 T^{-2}]$$[L] = [L]$$[T] = [T]$So,$[M^1 L^0 T^0] = [M^1 L^1 T^{-2}]^a [L]^b [T]^c$$[M^1 L^0 T^0] = [M^a L^{a+b} T^{-2a+c}]$Comparing the powers of $M, L,$ and $T$ on both sides:For $M$: $a = 1$For $L$: $a + b = 0 \implies 1 + b = 0 \implies b = -1$For $T$: $-2a + c = 0 \implies -2(1) + c = 0 \implies c = 2$Substituting these values back,we get the dimensional formula for mass as $M = F^1 L^{-1} T^2$ or $F L^{-1} T^2$.

If force $(F)$,length $(L)$,and time $(T)$ are assumed to be fundamental units,then the dimensional formula of mass will be:

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