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 Ω . Its internal resistance is
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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, then
where I 1 and I 2 are lengths of potentiometer wire with and without short circuited through a resistance.
Since, 
[ ∴ E = I (R + r) and V = IR]
∴ 
or 1 + 
or 
or r = 
× 10 = 1 Ω Ω
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