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The radius and height of a cylinder are measured as 5 cm and 10 cm respectively and there is an error of 0.02 cm in both the measurements . The relative error in volume is

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0.01

0.001

– 1

answer is A.

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Detailed Solution

$r = 5 h = 10 δr = δh = 0.02 v = {πr}^{2} h \log v = \log π + 2 \log x + \log h \frac{1}{v} δv = \frac{2}{r} δr + \frac{1}{h} δh = (\frac{2}{5} + \frac{1}{10}) (0.02) = (\frac{5}{10}) (0.02) = 0.01$

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