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 Omega`. Its internal resistance is

Options

A. 1
B. `0.5 Omega`
C. `2.0 Omega`
D. zero

Answer

A

Explanation

(a) 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
`(E)/(V) = (l_(1))/(l_(2))`
Where `l_(1)` and `l_(2)` are lengths of potendiometer wire with and without short circuited through a resistance.
Since `(E)/(V) = (R + r)/(R ) [:' E = I (R + r) "and" V = IR]`
`:. (R + r)/(R ) = (l_(1))/(l_(2))`
or `1 + (r )/(R ) = (110)/(100)` or `(r )/(R ) = (10)/(100)`
or `r = (1)/(10) xx 10 = 1 Omega`