Potentiometer wire of length `1 m` is connected in series with `490 Omega` resistance and `2 V` battery. If `0.2 mV/cm` is the potential gradient, then resistance of the potentiameter wire is approximately

Options

A. `4.9Omega`
B. `7.9Omega`
C. `5.9Omega`
D. `6.9 Omega`

Answer

A

Explanation

Potential drop across potentionmeter wire
`=(0.2xx10^-3)(100) = 0.02 V`
Now given resistance and potentiometer wire are in series with given battery. So potential will drop in direct ratio of resistance.
`:. 0.02/(2-0.02)=R/490`
`:. R=4.9Omega`