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Q.

If the radius of sphere is 3 cm. If an error of 0.03 cm is made in measuring the radius of the sphere, then the percentage error in its surface area is

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a

2

b

0.0002

c

0.02

d

0.002

answer is A.

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Detailed Solution

$r = 3 σr = 0.03 S = 4 π r^{2} \Rightarrow \log S = \log 4 π + \log r^{2} \frac{1}{S} σs = 2 \frac{1}{r} σr \frac{σS}{S} \times 100 = 2 \frac{σr}{r} \times 100 = 2 \times \frac{0.03}{3} \times 100 = 2$

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