Time (T),velocity (C),and angular momentum (h | vedclass

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Question

(A) Let mass be related to the fundamental quantities as $M \propto T^x C^y h^z$.The dimensional formula for mass is $[M^1 L^0 T^0]$.The dimensional f

Vedclass · Accepted Answer

$[M] = [T^{-1} C^{-2} h]$

Vedclass · Answer

(A) Let mass be related to the fundamental quantities as $M \propto T^x C^y h^z$.The dimensional formula for mass is $[M^1 L^0 T^0]$.The dimensional formulas for the given quantities are: $[T] = [T]$,$[C] = [L T^{-1}]$,and $[h] = [M L^2 T^{-1}]$.Substituting these into the proportionality equation:$[M^1 L^0 T^0] = [T]^x [L T^{-1}]^y [M L^2 T^{-1}]^z$$[M^1 L^0 T^0] = [M^z] [L^{y+2z}] [T^{x-y-z}]$Comparing the powers of $M, L,$ and $T$ on both sides:For $M$: $z = 1$For $L$: $y + 2z = 0 \implies y + 2(1) = 0 \implies y = -2$For $T$: $x - y - z = 0 \implies x - (-2) - 1 = 0 \implies x + 1 = 0 \implies x = -1$Thus,the dimensions of mass are $[M] = [T^{-1} C^{-2} h^1]$.

Column-$I$	Column-$II$
$(1)$ Torque	$(a)$ $M^1L^1T^{-1}$
$(2)$ Angular momentum	$(b)$ $M^1L^2T^{-1}$
$(3)$ Linear momentum	$(c)$ $M^1L^2T^{-2}$

Time $(T)$,velocity $(C)$,and angular momentum $(h)$ are chosen as fundamental quantities instead of mass,length,and time. In terms of these,the dimensions of mass would be

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Match the physical quantities in Column-$I$ with their corresponding dimensional formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ Torque $(a)$ $M^1L^1T^{-1}$

$(2)$ Angular momentum $(b)$ $M^1L^2T^{-1}$

$(3)$ Linear momentum $(c)$ $M^1L^2T^{-2}$

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