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. `1.0Omega`q
B. `0.5Omega`
C. `2.0Omega`
D. zero

Answer

A

Explanation

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)andl_(2)` are lengths of potentiometer wrie with and without short circuited through a resistance.
Since, `(E)/(V)=(R+r)/(R)`
`[:'E=I(R+r)andV=IR]`
`:.(R+r)/(R)=(l_(1))/(l_(2))`
or `1+(r)/(R)=(110)/(100)`
or `(r)/(R)=(10)/(100)`
or `r=(1)/(10)xx10=1Omega`