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A potentiometer of 20 Omega has 10 wires...

A potentiometer of `20 Omega` has 10 wires each of 1 metre length and the total resistance to be connected to the driving battery of emf 2 volts to produce a potential drop of `1muV` per millimetre . (Graph sheet is not required ) .

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The correct Answer is:
2
The total length of the 10 wires `=1xx10=10m`
Potential drop `1mu=(10^(-6))/(10^(-3))V//m=10^(-3)V//m`

So ,potential drop across the whole wire
`=10^(-3)xx10=10^(-2)V`
The resistance of the whole wire = `20Omega`
`therefore` Current in the potentiometer wire
`=(10^(-2))/(20)=0.5xx10^(-3)A`
`therefore` Net resistance of the circuit `=(2)/(0.5xx10^(-3))=4000Omega`
`therefore` The resistance connected to the driving battery
`= 4000-20 = 3980Omega`
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