Question

The error in the measurement to radius of a sphere is 0.3 %. what is the possible error in its surface area ?

Answer 1

% error = a Δx/x + b Δy/y + c Δz/z

Δr ------> Error in radius
Given that,
Δr/r ×100 = 0.3 %

Surface area of sphere ( A )= 4πr²

here 4π is a constant
∴ error of A is
               2 × Δr/r × 100
we know, 
               Δr/r × 100 = 0.3 
∴ %  error of A = 2 × 0.3
                        = 0.6% //