If the relative errors in the base radius and | vedclass

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(C) Given that, the relative error of height is $\frac{\delta h}{h} = 0.02$. The relative error of radius is $\frac{\delta r}{r} = 0.02$. The volume o

Vedclass · Accepted Answer

$6$

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(C) Given that, the relative error of height is $\frac{\delta h}{h} = 0.02$. The relative error of radius is $\frac{\delta r}{r} = 0.02$. The volume of the cone is $V = \frac{1}{3} \pi r^2 h$. Taking the natural logarithm on both sides, we get $\ln V = \ln \left( \frac{1}{3} \pi r^2 h ight)$. $\ln V = \ln \left( \frac{\pi}{3} ight) + 2 \ln r + \ln h$. Differentiating both sides, we get $\frac{\delta V}{V} = 2 \left( \frac{\delta r}{r} ight) + \left( \frac{\delta h}{h} ight)$. Substituting the given values, $\frac{\delta V}{V} = 2(0.02) + 0.02 = 0.04 + 0.02 = 0.06$. The percentage error in volume is $\frac{\delta V}{V} 	imes 100 = 0.06 	imes 100 = 6 \%$. Therefore, the percentage error in the volume is $6$.

If the relative errors in the base radius and the height of a cone are same and equal to $0.02$, then the percentage error in the volume of that cone is

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