If the measured value of the radius is 1.5 % | vedclass

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(C) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.For small percentage changes,the relative error in volume is given by $\frac{\Delta V

Vedclass · Accepted Answer

$4.6$

Vedclass · Answer

(C) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.For small percentage changes,the relative error in volume is given by $\frac{\Delta V}{V} = 3 	imes \frac{\Delta r}{r}$.Given that the radius is $1.5 \%$ larger,we have $\frac{\Delta r}{r} = 0.015$.Therefore,the percentage error in volume is $\frac{\Delta V}{V} 	imes 100 = 3 	imes 1.5 \% = 4.5 \%$.Using the exact calculation: If $r' = 1.015r$,then $V' = \frac{4}{3} \pi (1.015r)^3 = (1.015)^3 V$.$(1.015)^3 = 1.045678$.The percentage error is $(1.045678 - 1) 	imes 100 = 4.5678 \%$.Rounding to one decimal place,we get $4.6 \%$.

If the measured value of the radius is $1.5 \%$ larger,the percentage error (correct to one decimal place) made in calculating the volume of a sphere is

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