Question

a sphere of radius 10cm is to be turned out on a lathe. If the radius is made 0.1cm too long, what is the precentage of error made in the surface area of the sphere

Answer 1

Answer:

The percentage error is 8%

Step-by-step explanation:

Given

$r = 10cm$

$\triangle r =0.1cm$

Required

% error in the surface area

The surface area of a sphere is:

$S = \4\pi r^2$

Differentiate

$dS \approx 8\pi r\ dr$

Rewrite as:

$\triangle S =8\pi r \triangle r$ --- This represents the relative error

The percentage error is then calculated as:

$\% Error = \frac{Relative\ Error}{Surface\ Area}$

$\% Error = \frac{\triangle S}{S} * 100\%$

$\% Error = \frac{8\pi r \triangle r}{\pi r^2} *100\%$

$\% Error = \frac{8 \triangle r}{r} * 100\%$

$\% Error = \frac{8 *0.1}{10} * 100\%$

$\% Error = \frac{0.8}{10} * 100\%$

$\% Error = 0.08 * 100\%$

$\% Error = 8\%$