If energy is given by U = frac{Asqrt{x}}{x^2 | vedclass

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(B) According to the principle of homogeneity of dimensions,only quantities with the same dimensions can be added or subtracted.In the denominator,$x^

Vedclass · Accepted Answer

$ML^{11/2}T^{-2}$

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(B) According to the principle of homogeneity of dimensions,only quantities with the same dimensions can be added or subtracted.In the denominator,$x^2$ is added to $B$,so $[B] = [x^2] = [L^2]$.The energy $U$ has dimensions $[M L^2 T^{-2}]$.The equation is $U = \frac{A\sqrt{x}}{x^2 + B}$.Substituting the dimensions: $[M L^2 T^{-2}] = \frac{[A][L^{1/2}]}{[L^2]}$.Solving for $[A]$: $[A] = [M L^2 T^{-2}] 	imes \frac{[L^2]}{[L^{1/2}]} = [M L^{2+2-1/2} T^{-2}] = [M L^{7/2} T^{-2}]$.Now,calculate the dimensions of $AB$: $[AB] = [A][B] = [M L^{7/2} T^{-2}] 	imes [L^2] = [M L^{7/2+2} T^{-2}] = [M L^{11/2} T^{-2}]$.

If energy is given by $U = \frac{A\sqrt{x}}{x^2 + B}$,then the dimensions of $AB$ are:

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