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