Correct Answer - A
`(PTR)/100 = p[(1+R/100)^(2)-1](10,000 xx 2 xx Rs)/(100)`
`=10,000[(1+(Rc)/100)^(2)-1]`
`200Rs=10,000[(Rc^(2))/100^(2)+(2Rc)/100]`
`200Rx-200Rc=R_(c)^(2)`
`200(R_(s)-R_(c))=R_(c)^(2)`.
`R_(s)-R_(c)=R_(c)^(2)/200`
`R_(s) = R_(c)^(2)/200 +R_(c)`.
Since `R_(s)` and `R_(c)` are integers and `R_(c)^(2)` must be the smallest integer, it should be a multiple of 200 and a perfect square.
`therefore R_(c)^(2)=400`
`rArr R_(c) = 20%`
`R_(s) = 400/200 +20`
`R_(s)=22%`
`therefore` Minimum difference between `R_(s)` and `R_(c)` is 2.