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(B) $1$. Viscous force $(F)$ is a type of force, and the dimensional formula for force is $[M^1 L^1 T^{-2}]$. Thus, $(a)$ matches with $(i)$.$2$. Acco

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$(a-i), (b-ii)$

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(B) $1$. Viscous force $(F)$ is a type of force, and the dimensional formula for force is $[M^1 L^1 T^{-2}]$. Thus, $(a)$ matches with $(i)$.$2$. According to Newton's law of viscosity, $F = \eta A \frac{dv}{dx}$, where $\eta$ is the coefficient of viscosity, $A$ is the area, and $\frac{dv}{dx}$ is the velocity gradient.$3$. Rearranging for $\eta$: $\eta = \frac{F}{A (dv/dx)}$.$4$. Substituting dimensions: $[\eta] = \frac{[M^1 L^1 T^{-2}]}{[L^2] [T^{-1}]} = [M^1 L^{-1} T^{-1}]$. Thus, $(b)$ matches with $(ii)$.

Column-$I$	Column-$II$
$(a)$ Viscous force	$(i)$ $[M^1 L^1 T^{-2}]$
$(b)$ Coefficient of viscosity	$(ii)$ $[M^1 L^{-1} T^{-1}]$
	$(iii)$ $[M^1 L^{-1} T^{-2}]$

Different physical quantities are given in Column-$I$ and their dimensional formulas are given in Column-$II$. Match them appropriately.

Column-$I$ Column-$II$

$(a)$ Viscous force $(i)$ $[M^1 L^1 T^{-2}]$

$(b)$ Coefficient of viscosity $(ii)$ $[M^1 L^{-1} T^{-1}]$

$(iii)$ $[M^1 L^{-1} T^{-2}]$

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Different physical quantities are given in Column-$I$ and their dimensional formulas are given in Column-$II$. Match them appropriately. Column-$I$ Column-$II$ $(a)$ Viscous force $(i)$ $[M^1 L^1 T^{-2}]$ $(b)$ Coefficient of viscosity $(ii)$ $[M^1 L^{-1} T^{-1}]$ $(iii)$ $[M^1 L^{-1} T^{-2}]$