If the term independent of $x$ in the expansion of $\left(\sqrt{\mathrm{ax}}{ }^2+\frac{1}{2 \mathrm{x}^3}\right)^{10}$ is 105 , then $\mathrm{a}^2$ is equal to :

  • [JEE MAIN 2024]
  • A

    $4$

  • B

    $9$

  • C

    $6$

  • D

    $2$

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