A potentiometer wire has a length of `10 m` and resistance `4 Omega m^(-1)`. An accumulator of emf `2 V` and a resistance box are connected in series with it. Culculate the resistance to be introduced in the box so as to get a potential gradient of
(a) `0.1 V//m` and (b) `0.1 mVm^(-1)`.

Answer

(a) `40 Omega` (b) `79960 Omega`

Explanation

Let `R `be the resistance to be introduced in the box. Current in the potentiomenter wire is given by
`1 = (E)/(R + l_(rho))`
where `rho` is the resistance per unit length of the wire and `l` is the length of the wire. Now, potential gradient is
`k = I rho = (E_(rho))/(R + l_(rho))`
Here, `l = 10 m, rho = 4 Omegam^(-1)`
(i) For `k = 0.1 Vm^(-1)`, we have
`0.1 = (2xx4)/(R + 10 xx 4) = (8)/(R + 40)` or `R = (4)/(0.1) = 40 Omega`
(ii) For `k = 0.1 mVm^(-1) = 0.1 xx 10^(-3) Vm^(-1)` , we have
`0.1 xx 10^(-3) = (2 xx 4)/(R + 10 xx 4)`
or `10^(-4) = (8)/(R + 40)` or `R = 79960 Omega`