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. 2V, so as to get a potential gradient 1 mV per cm on the wire is :

Options

A. `32 Omega`
B. `40 Omega`
C. `44 Omega`
D. `48 Omega`

Answer

A

Explanation

`32 Omega`
Potential gradient, `sigma = (1mV)/(cm)`
= ` (10^(-3))/(10^(-2) m) =0.1 (V)/(m)`
Suppose, resistance R is connected in series with potentiometer.
`therefore ` Current in circuit I = `(V)/(8 + R) = (2)/(8 + R) `
Potential gradient , `sigma = I rho `
`0.1 = (2)/(8+ R) xx (8)/(4)`
` [ because gamma = ("resistance of wire")/("length of wire") ] `
8 + R = `(4)/(1)`
`therefore R = 40 - 8 `
`therefore R = 32 Omega`