If the radius of a sphere is $(5.3 \pm 0.1) \; cm$,then the 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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