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

The percentage error is 8%

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 = (Relative\ Error)/(Surface\ Area)`

`\% Error = (\triangle S)/(S) * 100\%`

`\% Error = (8\pi r \triangle r)/(\pi r^2) *100\%`

`\% Error = (8 \triangle r)/(r) * 100\%`

`\% Error = (8 *0.1)/(10) * 100\%`

`\% Error = (0.8)/(10) * 100\%`

`\% Error = 0.08 * 100\%`

`\% Error = 8\%`