The correct option is D 1 %
The percentage error in measurement of radius is given, Δrr×100=0.5%
Thus, Δrr=0.5/100=0.005
The surface area of sphere is S=4πr2
Take ln and differentiate, ΔSS=2Δrr=2×0.005=0.01
The permissible or % error in the measurement of surface area =ΔSS×100=0.01×100=1%