If the error in measuring the radius of the sphere is 2% and that in measuring its mass is 3%, Then the error in measuring the density of material of the sphere is:

Options

A. 0.05
B. 0.07
C. 0.09
D. 0.11

Answer

C

Explanation

As density `=("mass")/("volume")`
`thereforerho=(M)/((4)/(3)piR^(3))=(3)/(4)(M)/(piR^(3))`
`therefore` The percentage error in density is
`(Deltarho)/(rho)xx100%=((DeltaM)/(M)+3(DeltaR)/(R))xx100%`
`=3%+3(2%)=3%+6%=9%`