Question

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

Answer 1

% error = a Δx/x + b Δy/y + c Δz/z

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

Surface are 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% //

Answer 2

surface area of a sphere= 4π r^2
 
while considering the percentage error , it will be like ,
 4 π * 2*Δr/r  *100
 
in the question its given that  Δr/r *100 =0.3
4π is a constant 


substituting the given value in the equation  we get

2*0.3= 0.6 % is the possible error in the surface area