If there is 2% error in measuring the radius of sphere, then…………… will be the percentage error in the surface area.

Options

A. 0.03
B. 0.01
C. 0.04
D. 0.02

Answer

C

Explanation

Given `(Deltar)/r xx100=2impliesDeltar=(2r)/100`
Surfasce area of sphere `S=4pir^(2)`
`DeltaS=4pi2r.Deltar=8pir.(2r)/100=(16pi r^(2))/100`
Now, percentage error in surface area
`=(Deltas)/Sxx100=(16pir^(2))/100xx1/(4pir^(2))xx100=4%`