If the radius of a sphere is measured as 9 c | vedclass

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

Question

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.Given, $r = 9 \ cm$ and $\Delta r = 0.03 \ cm$.The surface

Vedclass · Accepted Answer

$2.16$

Vedclass · Answer

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.Given, $r = 9 \ cm$ and $\Delta r = 0.03 \ cm$.The surface area $S$ of a sphere is given by $S = 4 \pi r^2$.To find the approximate error in the surface area, we differentiate $S$ with respect to $r$:$\frac{dS}{dr} = 8 \pi r$.Using the differential approximation, $\Delta S \approx \frac{dS}{dr} 	imes \Delta r$.Substituting the values:$\Delta S = 8 \pi 	imes 9 	imes 0.03$.$\Delta S = 72 \pi 	imes 0.03 = 2.16 \pi \ cm^2$.Thus, the approximate error in calculating the surface area is $2.16 \pi \ cm^2$.

If the radius of a sphere is measured as $9 \ cm$ with an error of $0.03 \ cm$, then find the approximate error in calculating its surface area. (in $\pi \ cm^2$)

Similar Questions

The semi-vertical angle of a right circular cone is $30^{\circ}$. If the height of the cone is $6.125 \text{ cm}$,then the approximate value of the volume of the cone (in cubic $\text{cm}$) is

The time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$. If the length is increased by $2 \%$,then the approximate change in the time period is:

The approximate value of $\tan ^{-1}(0.999)$ is (use $\pi=3.1415$ )

Find the approximate value of $(0.009)^{\frac{1}{3}}$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module