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Question

(A) The potential energy $U$ has dimensions of work or energy,which is $[ML^2T^{-2}]$.According to the principle of homogeneity of dimensions,only qua

Vedclass · Accepted Answer

$[ML^{5/2}T^{-2}], [L]$

Vedclass · Answer

(A) The potential energy $U$ has dimensions of work or energy,which is $[ML^2T^{-2}]$.According to the principle of homogeneity of dimensions,only quantities with the same dimensions can be added or subtracted. In the denominator,$x + B$,$x$ is a distance,so $B$ must also have the dimensions of length.Therefore,$[B] = [L]$.Now,substitute the dimensions into the equation:$[ML^2T^{-2}] = \frac{[A] \cdot [L^{1/2}]}{[L]}$$[ML^2T^{-2}] = [A] \cdot [L^{-1/2}]$$[A] = [ML^2T^{-2}] \cdot [L^{1/2}]$$[A] = [ML^{5/2}T^{-2}]$Thus,the dimensions of $A$ and $B$ are $[ML^{5/2}T^{-2}]$ and $[L]$ respectively. The correct option is $A$.

The potential energy $U$ of a particle varies with distance $x$ from a fixed origin as $U = \frac{A \sqrt{x}}{x + B}$,where $A$ and $B$ are constants. The dimensions of $A$ and $B$ are respectively

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