Question

If error in measuring radius of a sphere is 2% then what will be percentage error in measuring its (i) surface area (ii) volume?

Answer 1

Answer:

Percentage error in radius is given as 2% i.e.

r

Δr

×100=2 %

Volume of sphere V=

3

4π

r

3

Percentage error in volume

V

ΔV

×100=3×

r

Δr

×100=3×2=6 % .

Explanation:

volume