If dimensions of critical velocity v_c of a l | vedclass

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(B) Given the dimensional relation: $[v_c] = [\eta^x \rho^y r^z]$ $(i)$Writing the dimensions of the physical quantities:$[v_c] = [M^0 L T^{-1}]$$[\et

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$1, -1, -1$

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(B) Given the dimensional relation: $[v_c] = [\eta^x \rho^y r^z]$ $(i)$Writing the dimensions of the physical quantities:$[v_c] = [M^0 L T^{-1}]$$[\eta] = [M L^{-1} T^{-1}]$$[\rho] = [M L^{-3} T^0]$$[r] = [M^0 L T^0]$Substituting these into equation $(i)$:$[M^0 L T^{-1}] = [M L^{-1} T^{-1}]^x [M L^{-3} T^0]^y [M^0 L T^0]^z$$[M^0 L T^{-1}] = [M^{x+y} L^{-x-3y+z} T^{-x}]$Applying the principle of homogeneity of dimensions,we equate the powers of $M, L,$ and $T$ on both sides:$x + y = 0$ $(ii)$$-x - 3y + z = 1$ $(iii)$$-x = -1$ $(iv)$From $(iv)$,we get $x = 1$.Substituting $x = 1$ into $(ii)$,we get $1 + y = 0$,so $y = -1$.Substituting $x = 1$ and $y = -1$ into $(iii)$,we get $-1 - 3(-1) + z = 1 \implies -1 + 3 + z = 1 \implies 2 + z = 1 \implies z = -1$.Thus,the values are $x = 1, y = -1, z = -1$.

If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as $[\eta^x \rho^y r^z]$,where $\eta, \rho,$ and $r$ are the coefficient of viscosity of the liquid,density of the liquid,and radius of the tube respectively,then the values of $x, y,$ and $z$ are given by:

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