A) \[1.0\Omega \]
B) \[0.5\,\Omega \]
C) \[2.0\,\Omega \]
D) zero
Correct Answer: A
Solution :
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 \[\frac{E}{V}=\frac{{{l}_{1}}}{{{l}_{2}}}\] where \[{{l}_{1}}\] and \[{{l}_{2}}\] are lengths of potentiometer wire with and without short circuited through a resistance. Since \[\frac{E}{V}=\frac{R+r}{R}\]\[[\because \,E=I(R+r)\,and\,V=IR]\] \[\therefore \] \[\frac{R+r}{R}=\frac{{{l}_{1}}}{{{l}_{2}}}\] or \[1+\frac{r}{R}=\frac{110}{100}\]or \[\frac{r}{R}=\frac{10}{100}\] or \[r=\frac{1}{10}\times 10=1\,\Omega \]
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