The potential energy of a particle varies with distance x from a fixed origin as V=Axx+B where A and B are constants. The dimensions of AB are

V=AXX+B∴B=[L]V=ML2 T-2∴ML2 T-2=A[L]12 L⇒ML3 T-2=A[L]1/2A=ML3 T-2L12=ML3 L-12 T-2=ML52 T-2∴AB=ML52 T-2[L]=ML72 T-2

The potential energy of a particle varies with distance x from a fixed origin as V=Axx+B where A and B are constants. The dimensions of AB are

V=AXX+B∴B=[L]V=ML2 T-2∴ML2 T-2=A[L]12 L⇒ML3 T-2=A[L]1/2A=ML3 T-2L12=ML3 L-12 T-2=ML52 T-2∴AB=ML52 T-2[L]=ML72 T-2

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

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$M^{1} L^{5 / 2} T^{- 2}$

$M^{1} L^{2} T^{- 2}$

$M^{3 / 2} L^{5 / 2} T^{- 2}$

$M^{1} L^{7 / 2} T^{- 2}$

answer is D.

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Detailed Solution

$\begin{array}{rcl} V & = & \frac{A \sqrt{X}}{X + B} \\ ∴ B & = & [L] \\ V & = & [{ML}^{2} T^{- 2}] \\ ∴ [{ML}^{2} T^{- 2}] & = & \frac{A [L]^{\frac{1}{2}}}{L} \Rightarrow [{ML}^{3} T^{- 2}] = A [L]^{1 / 2} \\ A & = & \frac{[{ML}^{3} T^{- 2}]}{L^{\frac{1}{2}}} = {ML}^{3} L^{\frac{- 1}{2}} T^{- 2} = [{ML}^{\frac{5}{2}} T^{- 2}] \\ ∴ AB & = & [{ML}^{\frac{5}{2}} T^{- 2}] [L] = {ML}^{\frac{7}{2}} T^{- 2} \end{array}$