The count rate from `100cm^(3)` of a radioactive liquid is `c`. Some of this liquid is now discarded. The count rate of the remaining liquid is found to be `c//10` after three half-lives. The volume of the remaining liquid, in `cm^(3)`, is

Options

A. `20`
B. `40`
C. `60`
D. `80`

Answer

D

Explanation

Initial activity/volume `= (c )/(100)`
Let `V:` volume of remaining liquid
Activity after three half-lives `= (c//10)/(V)`
`A = A_(0)((1)/(2))^(n)`
`(c//10)/(V) = (c )/(100) ((1)/(2))^(3)`
`V = (c//10)/(c//800) = 80cm^(3)`