Correct Answer - B
To get maximum equivalent resistance all resistances must be connected in series
`therefore (R_("eq"))_("max")=R+R+R...n"times"=nR`
To get minimum equivalent resistance all resistances myst be connected in parallel.
`therefore (1)/(R_("eq"))_("min")=(1)/(R)+(1)/(R)+....n"time", (1)/(R_("eq"))_("min")=(n)/(R)`
`(R_("eq"))_("min")=(R)/(n)therefore (R_("eq"))_("max")/(R_("eq"))_("min")=(nR)/(R//n)=n^(2)`