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The potential energy of a particle moving along \(X\)-direction varies as \(V=\frac{A{x}^{2}}{\sqrt{x}+B}\). where x is …

Q1

The potential energy of a particle moving along \(X\)-direction varies as \(V=\frac{A{x}^{2}}{\sqrt{x}+B}\). where x is distance, The dimensions of \(\frac{{A}^{2}}{B}\) are:

a

\(\left[{\mathrm{M}}^{3/2}{\mathrm{L}}^{1/2}{\mathrm{T}}^{-3}\right]\)

b

\(\left[{\mathrm{M}}^{1/2}{\mathrm{LT}}^{-3}\right]\)

c

\(\left[{\mathrm{M}}^{2}{\mathrm{L}}^{1/2}{\mathrm{T}}^{-4}\right]\)

d

\(\left[{\mathrm{ML}}^{2}{\mathrm{T}}^{-4}\right]\)

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