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Q.

If there is an error of 310% in the volume of a sphere then the percentage error in its radius is

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a

210

b

310

c

110

d

3

answer is A.

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Detailed Solution

δvv×100=310 v=43π r3  ⇒  log v=log 43π+3log r 1v δv=3 1r δr δrr×100=3 δrr×100  ⇒  δrr×100=110

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