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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $cm^3$)

Similar Questions

Use differentials to approximate $\sqrt{36.6}$.

If the radius of a circle increases from $3 \, cm$ to $3.2 \, cm$,then the increase in the area of the circle is

The approximate value of $\sin 31^{\circ}$ is

Find the approximate value of $(1.999)^{5}$.

The radius of a cone of height $9 \text{ units}$ is changed from $2 \text{ units}$ to $2.12 \text{ units}$. The exact change and approximate change in the volume of the cone are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module