Question

the radius of the sphere is 20+_2 the percentage error in its surface area is​

Answer 1

Radius = 20 ± 2 m

relative error (RE) = ΔR / R

ΔR = ±2 m

R = 20 m

RE = 2/20 = 0.1 m

Now,

surface area (SA) = 4πR²

thus

ΔSA / SA = 2 ΔR/R

for percentage error multiply both the sides by 100

%SA = 2 %R

%SA = 2 (0.1) x 100 = 20 %

Thus the percentage error in the surface area of the sphere = 20%