A potentiometer wire of length 100 cm has a resistance of `100Omega` it is connected in series with a resistance and a battery of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. what is the value of the external resistance?

Options

A. `790Omega`
B. `890Omega`
C. `990Omega`
D. `1090Omega`

Answer

A

Explanation

The current in the potentiometer wire AC is `I=(2)/(10+R)`
The potential difference across the potentiometer wire is V = current `xx` resistance = `(2)/(10+R)xx10`
The length of the wire is `l=100cm`.
So, the potential gradient along the wire is
`k=(V)/(l)=((2)/(10+R))xx(10)/(100)" "...(i)`
The source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire i.e. `10xx10^(-3)=kxx40`
or `10xx10^(-3)=(2)/((10+R))xx(40)/(10)` (Using (i))
or `R=790Omega`