If Surface tension (S),Moment of Inertia (I) | vedclass

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(A) Let the linear momentum $P$ be expressed as $P = k S^a I^b h^c$,where $k$ is a dimensionless constant.Dimensional formulas are:$P = [MLT^{-1}]$$S

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$S^{1/2} I^{1/2} h^0$

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(A) Let the linear momentum $P$ be expressed as $P = k S^a I^b h^c$,where $k$ is a dimensionless constant.Dimensional formulas are:$P = [MLT^{-1}]$$S = [MT^{-2}]$$I = [ML^2]$$h = [ML^2T^{-1}]$Substituting these into the equation:$[MLT^{-1}] = [MT^{-2}]^a [ML^2]^b [ML^2T^{-1}]^c$$[MLT^{-1}] = M^{a+b+c} L^{2b+2c} T^{-2a-c}$Comparing the powers of $M, L,$ and $T$ on both sides:$a + b + c = 1$ $(1)$$2b + 2c = 1$ $(2)$$-2a - c = -1$ $(3)$From $(2)$,$b + c = 1/2$. Substituting this into $(1)$:$a + 1/2 = 1 \implies a = 1/2$From $(3)$,$c = 1 - 2a = 1 - 2(1/2) = 0$.Substituting $c = 0$ into $(2)$,$2b = 1 \implies b = 1/2$.Thus,the dimensional formula for linear momentum is $S^{1/2} I^{1/2} h^0$.

If Surface tension $(S)$,Moment of Inertia $(I)$ and Planck's constant $(h)$ were to be taken as the fundamental units,the dimensional formula for linear momentum would be

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