The driver cell of a potentiometer has an emf of 2V and negligible internal resistance. The potentiometer wire has a resistance of 5$\Omega$ and is 1m long. The resistance which must be connected in series with the wire so as to have a potential difference of 5mV across the whole wire is
Answer & explanation
Correct answer: option 3
In order to have a potential drop of 5 mV , across a wire of resistance 5 W, the current flowing in the wire should be:
$ I = \frac{5\times 10^{-3}}{5} = 10^{-3}A$
If R is the resistance to be connected in series with the wire, then $\frac{2}{R+5} = 10^{-3}A$
which gives $R = 1995 \Omega$