If the error in measurement of radius of a sphere is 2% then the error in the determination of volume of the sphere will be

Options

A. `4%`
B. `6%`
C. `8%`
D. `2%`

Answer

B

Explanation

Let the actual and the incorrect radii be r and r respectively.
According to the question,
` (r )/( r) =(102)/(100) =1.02`
[We could also take `( r)/(r )=(98)/(100) =0.98`]
`:. ("incorrect volume")/( "actual volume") =((4)/(3)pir^(3))/((4)/(3)pir^(3))=((r)/(r))^(3)`
`=(1.02)^(3) ~~1.06`
Therefore percentage error =`(1.06-1)xx100%=6%`
Alternate method:
Volume V`=(4)/(3) pir^(3) " ""or", In V =In ((4)/(3)pi)+3` In r
Differentiating we get `(dV)/(V) =0 +3(dr)/(r)`
Here error in measuring radius `(dr)/(r) = 2%`
Therefore error in volume measurement
`(dV)/(V) = 3 xx2% = 6%`
[This method is applicable only when percentage error <<100]