1.Units, Dimensions and Measurement
easy

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

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

Solution

(c) Volume of sphere $(V) = \frac{4}{3}\pi {r^3}$

$\%$ error in volume $ = 3 \times \frac{{\Delta r}}{r} \times 100$$ = \left( {3 \times \frac{{0.1}}{{5.3}}} \right) \times 100$

Standard 11
Physics

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