Q135 / 200 · NEET 2024 (Re-NEET) · Physics

The potential energy of a particle moving along $x$-direction varies as $V=\frac{A x^2}{\sqrt{x}+B}$. The dimensions of $\frac{A^2}{B}$ are:

  • A
    $\left[\mathrm{M}^{3 / 2} \mathrm{~L}^{1 / 2} \mathrm{~T}^{-3}\right]$
  • B
    $\left[M^{1 / 2} L^{-3}\right]$
  • Correct answer
    C
    $\left[\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}\right]$
  • D
    $\left[\mathrm{ML}^2 \mathrm{~T}^{-4}\right]$

Solution

$V=\frac{A x^2}{\sqrt{x}+B}$ As per homogeneous rule $B=\sqrt{L}$ $\mathrm{ML}^2 \mathrm{~T}^{-2}=\frac{A \mathrm{~L}^2}{\mathrm{~L}^{1 / 2}}$ $A=\mathrm{ML}^{1 / 2} \mathrm{~T}^{-2}$ $\frac{A^2}{B}=\frac{\mathrm{M}^2 L T^{-4}}{\mathrm{~L}^{1 / 2}}=\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}$

Hence, the correct answer is option: 'C'.

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