A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively, with and without being short circuited through a resistance of 10  Ω Its internal resistance is

This problem is based on the application of potentiometer in which we find the internal resistance of a cell.In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential difference, thenEV=l1l2where l1 and l2 are lengths of potentiometer wire with and without short circuited through a resistance.Since  EV  = R+rR   ∵   E=IR+r  and  V=IR∴      R+rR  =  l1l2or      1+rR  =  110100      or    rR=10100or    r=110  ×  10=1  Ω