Quick practice

Question 1 of 5

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

A

\left( {\frac{1}{3}} \right) \times 0.01 \times \left( {\frac{{100}}{{5.3}}} \right)

B

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

C

3 + 6.01 \times \left( {\frac{{100}}{{5.3}}} \right)

D

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

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