A potentiometer wire has length `4 m` and resistance `8 Omega`. The resistance that must be connected in series with the wire and an accumulator of e.m.f. `2 V`, so as the get a potential gradient `1 mV` per cm` on the wire is

Options

A. `40 Omega`
B. `44 Omega`
C. `48 Omega`
D. ` 52 Omega`

Answer

d

Explanation

Potential gradent `= (1mV)/(1cm) = 0.1V//m`
Let the resistance to be connected is `R` then
`1 = (2)/(8+R)`
Potential drop across the potentiometer in `R` wire
`= (8xx2)/(8+R) = (16)/(8+R)`
Potential gradent
`= ((16)/(8+R)) xx(1)/(4) V//m = (4)/(8+R) = 0.1`
`R = 32 Omega`