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

Explanation

As shown in the figure, if R is the unknown resistance, the current in the circuit
`I=(V)/(r+R)=(2)/(10+R)`
Now, as the 100 cm wire has a resistance of `10Omega`, the resistance of 40 cm of wire will be `40 xx (10//100) = 4 ohm`. Potential drop across 40 cm wire will be` V = I xx 4` But, here `V = 10 mV `(given)
`i=(2)/((10+R))=(1)/(400),i.e.R=790Omega`