If text{force} = frac{alpha}{text{density} + | vedclass

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(B) The given equation is $	ext{Force} = \frac{\alpha}{	ext{density} + \beta^3}$.According to the principle of homogeneity of dimensions,quantities

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$[M^2 L^{-2} T^{-2}], [M^{1/3} L^{-1} T^0]$

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(B) The given equation is $	ext{Force} = \frac{\alpha}{	ext{density} + \beta^3}$.According to the principle of homogeneity of dimensions,quantities added together must have the same dimensions.Therefore,the dimension of $\beta^3$ must be equal to the dimension of density.Dimension of density = $[M L^{-3} T^0]$.So,$[\beta^3] = [M L^{-3} T^0]$.Taking the cube root,$[\beta] = [M^{1/3} L^{-1} T^0]$.Now,the dimension of force is $[M L T^{-2}]$.From the equation,$\alpha = 	ext{Force} 	imes (	ext{density} + \beta^3)$.Since $(	ext{density} + \beta^3)$ has the same dimension as density,$[\alpha] = [M L T^{-2}] 	imes [M L^{-3} T^0] = [M^2 L^{-2} T^{-2}]$.Thus,the dimensions are $[M^2 L^{-2} T^{-2}]$ and $[M^{1/3} L^{-1} T^0]$.

List-$I$	List-$II$
$A$. Meter $(L)$	$I$. $\sqrt{\frac{hc}{G}}$
$B$. Second $(S)$	$II$. $\sqrt{\frac{Gh}{c^5}}$
$C$. Kilogram $(M)$	$III$. $\sqrt{\frac{L^2c^3}{Gh}}$
$D$. Kelvin $(K)$	$IV$. $\sqrt{\frac{Gh}{c^3}}$

If $\text{force} = \frac{\alpha}{\text{density} + \beta^3}$,then the dimensional formulae of $\alpha$ and $\beta$ are respectively:

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