If the percentage error in the radius of a sphere is `alpha` , find the percentage error in its volume.

Options

A. `3/2 a %`
B. `2/3 a %`
C. `3 a %`
D. none of these

Answer

A

Explanation

Let r the radius, S the surface area the V the volume of the sphere. Then, `S= 4pi^(2) and V=4/3pi r^(3)`
Let `Delta r, Delta S and Delta V` be the errors in r, S and V respectively. Then, Let `Delta r, Delta S and Delta V` be the errors in r, S and V respectively. Then,
`Delta S=(dS)/(dr) Delta r and Delta V = (dV)/(dr) Delta r`
`rArrDelta S=8pir" "Deltar and Delta V= 4pir^(2)" "Deltar`
`rArr" "(DeltaS)/S= (8pir)/(4pir^(2)) Deltar and (DeltaV)/V=(4pir^(2)" "Deltar)/(4/3pir^(3))`
`rArr (Delta S)/S xx100=2((Deltar)/r xx100)and (Delta V)/V xx100=3((Delta r)/r xx100)`
`rArr" "a=2((Delta r)/r xx100) and (DeltaV)/Vxx100=3((Delta r)/r xx100)" "[:'(Delta S)/S xx100=a(given)]`
`rArr (Delta V)/V xx100 = 3(a/2)=3/2a`