Let [{varepsilon _0}] denote the dimensional | vedclass

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(B) From Coulomb's law,the force between two charges is given by $F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{R^2}}}$.Rearranging for p

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$[ {\varepsilon _0} ]=[M^{-1}L^{-3}T^4A^2]$

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(B) From Coulomb's law,the force between two charges is given by $F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{R^2}}}$.Rearranging for permittivity,we get ${\varepsilon _0} = \frac{{{q_1}{q_2}}}{{4\pi F{R^2}}}$.The dimensional formula for charge $q$ is $[AT]$,for force $F$ is $[MLT^{-2}]$,and for distance $R$ is $[L]$.Substituting these into the formula: $[{\varepsilon _0}] = \frac{{[AT][AT]}}{{[MLT^{-2}][L^2]}} = \frac{{[A^2T^2]}}{{[ML^3T^{-2}]}}$.Simplifying the expression,we get $[{\varepsilon _0}] = [M^{-1}L^{-3}T^4A^2]$.

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

Let $[{\varepsilon _0}]$ denote the dimensional formula of the permittivity of vacuum. If $M =$ mass,$L =$ length,$T =$ time and $A =$ electric current,then:

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

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

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