There is a possible error of 0.02 cm in meas | vedclass

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(C) Let the diameter be $D = 14 \ cm$, so the radius $r = 7 \ cm$. The error in diameter is $\Delta D = 0.02 \ cm$, so the error in radius is $\Delta

Vedclass · Accepted Answer

$1.54$

Vedclass · Answer

(C) Let the diameter be $D = 14 \ cm$, so the radius $r = 7 \ cm$. The error in diameter is $\Delta D = 0.02 \ cm$, so the error in radius is $\Delta r = \frac{\Delta D}{2} = 0.01 \ cm$. Given the semi-vertical angle $\alpha = 45^{\circ}$, the height $h$ of the cone is $h = \frac{r}{	an(\alpha)} = \frac{r}{	an(45^{\circ})} = r$. The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^3$. Differentiating with respect to $r$, we get $\frac{dV}{dr} = \pi r^2$. The approximate error in volume is $\Delta V \approx \frac{dV}{dr} 	imes \Delta r$. Substituting the values, $\Delta V \approx \pi 	imes (7)^2 	imes 0.01 = 49 \pi 	imes 0.01 = 0.49 \pi$. Using $\pi \approx \frac{22}{7}$, we get $\Delta V \approx 0.49 	imes \frac{22}{7} = 0.07 	imes 22 = 1.54 \ cm^3$.

There is a possible error of $0.02 \ cm$ in measuring the base diameter of a right circular cone as $14 \ cm$. If the semi-vertical angle of the cone is $45^{\circ}$, then the approximate error in its volume is (in $cm^3$)

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