The potential energy u of a particle varies with distance x from a fixed origin as u=(A √x/x+B), where A and B are constants. The dimensions of A and B are respectively

Question

The potential energy u of a particle varies with distance x from a fixed origin as u=(A √x/x+B), where A and B are constants. The dimensions of A and B are respectively (A) [ ML .5 / 2 T -2],[ L ]. (B) [ MLT -2],[ L 2] (C) [ L ],[ ML 3 / 2 T -2] (D) [ L 2],[ MLT -2]

Tardigrade Admin · Accepted Answer

u=(A √x/x+B) By the principle of homogeneity, x=B (dimensionally) ⇒ B=[ L ] and [ ML 2 T -2]=(A L1 / 2/L) [ ML 2 T -2]=A L-1 / 2 A=[ ML 3 / 2 T -2]

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

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

$[M L T^{- 2}], [L^{2}]$

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

$[L^{2}], [M L T^{- 2}]$

$u = \frac{A x}{x + B}$
By the principle of homogeneity, $x = B$ (dimensionally)
$\Rightarrow B = [L]$
and $[M L^{2} T^{- 2}] = \frac{A L ^{1/2}}{L}$
$[M L^{2} T^{- 2}] = A L^{- 1/2}$
$A = [M L^{3/2} T^{- 2}]$

Q. The potential energy u of a particle varies with distance x from a fixed origin as u=x+BAx​​, where A and B are constants. The dimensions of A and B are respectively

[ML5/2T−2],[L]

[MLT−2],[L2]

[L],[ML3/2T−2]

[L2],[MLT−2]