There is a potentiometer wire of length 1200 cm and a 60 mA curreut is flowing in it. A battery of emf 5 V and internal resistance of 20 `Omega` is balanced on this potentiometer wfre with a balancing length 1000 cm. The resistance of the potentiometer wire is

Options

A. `60 Omega`
B. `80 Omega`
C. `1000 Omega`
D. `120 Omega`

Answer

C

Explanation

`100 Omega`

Let terminal voltage of primary battery is `V_(p)`. For any neutral point on potentiometer wire, `V_(p)` remains constant .
`therefore ` potential gradient,
`(V)/(l) = (V_(p))/(L)`
`therefore (5)/(1000) = (V_(p))/(1200)`
`therefore = (1200 xx 5)/(1000) = 6V`
`therefore ` Resistance of potentiometer wire
`R_(p) = (V_(p))/(I)`
`therefore R_(p) = (6)/(60 xx 10^(-3)) `
`therefore R_(p) =100 Omega`