if Energy is given by U = frac{{Asqrt x }}{{{ | vedclass

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Question

$[{x^2}]\, = [B]$ $[B] = [L^2]$

$\left[ U \right] = \frac{{[A]\,[{x^{1/2}}]}}{{[{x^2}]\, + \,[B]}}$

$&rArr; [M{L^2}{T^{ - 2}}] = \frac{{[A]\,[{L

Vedclass · Accepted Answer

$M{L^{11/2}}{T^{ - 2}}$

Vedclass · Answer

$[{x^2}]\, = [B]$ $[B] = [L^2]$

$\left[ U ight] = \frac{{[A]\,[{x^{1/2}}]}}{{[{x^2}]\, + \,[B]}}$

$&rArr; [M{L^2}{T^{ - 2}}] = \frac{{[A]\,[{L^{1/2}}]}}{{[{L^2}]}}$

$	herefore\,[A] = [M{L^{7/2}}{T^{ - 2}}]$

$[AB]\, = \,[M{L^{7/2}}{T^{ - 2}}] 	imes [{L^2}]$ $ = [M{L^{11/2}}{T^{ - 2}}]$

if Energy is given by $U = \frac{{A\sqrt x }}{{{x^2} + B}},\,$, then dimensions of $AB$ is

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