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. zero
B. `6.5 Omga`
C. `0.7 Omega`
D. `1 Omega`

Answer

D

Explanation

The internal resistance of a cell r
`= R [(l_(1)-l_(2))/(l_(2))]=10[(110-100)/(100)]`
`= 10xx(1)/(10)=1 Omega`
where `l_(1) and l_(2)` are the balancinglengths with and without being short circuited.