Force (F) and density (d) are related as F = | vedclass

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(A) According to the principle of homogeneity of dimensions,the dimensions of terms added together must be the same. Therefore,the dimension of $\beta

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) According to the principle of homogeneity of dimensions,the dimensions of terms added together must be the same. Therefore,the dimension of $\beta$ must be equal to the dimension of $\sqrt{d}$.$[\beta] = [d]^{1/2} = [M^1 L^{-3} T^0]^{1/2} = [M^{1/2} L^{-3/2} T^0]$.Now,for the force equation: $F = \frac{\alpha}{\beta + \sqrt{d}}$.The dimension of the denominator $(\beta + \sqrt{d})$ is the same as $[d]^{1/2} = [M^{1/2} L^{-3/2} T^0]$.Thus,$[F] = \frac{[\alpha]}{[M^{1/2} L^{-3/2} T^0]}$.$[\alpha] = [F] 	imes [M^{1/2} L^{-3/2} T^0] = [M^1 L^1 T^{-2}] 	imes [M^{1/2} L^{-3/2} T^0] = [M^{3/2} L^{-1/2} T^{-2}]$.Comparing these results with the given options,option $A$ is correct.

Force $(F)$ and density $(d)$ are related as $F = \frac{\alpha}{\beta + \sqrt{d}}$. The dimensions of $\alpha$ and $\beta$ are:

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