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Q.

There is an error of $\pm$ 0.04 cm in the measurment of the diameter of a sphre. When the radius is 10cm, the percntage error in the volumeof the sphere is

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a

$\pm 12$

b

$\pm 1.0$

c

$\pm 10.8$

d

$\pm 0.6$

answer is D.

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Detailed Solution

$δ (2 r) = \pm 0.04 \Rightarrow δ r = \pm 0.02 r = 10 Volume v = \frac{4}{3} π r^{3} \log v = \log (\frac{4}{3} π) + \log r^{3} \frac{1}{v} δ v = 3 \frac{1}{r} δ r \Rightarrow \frac{1}{v} δ v \times 100 = \frac{3 \times 0.02}{10} \times 100 = 0.6$

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