If radius of the sphere is $(5.3 \pm 0.1)\;cm$. Then percentage error in its volume will be

  • A

    $3 + 6.01 \times \frac{{100}}{{5.3}}$

  • B

    $\frac{1}{3} \times 0.01 \times \frac{{100}}{{5.3}}$

  • C

    $\left( {\frac{{3 \times 0.1}}{{5.3}}} \right) \times 100$

  • D

    $\frac{{0.1}}{{5.3}} \times 100$

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