A 10 m long wire ofresistance `20 Omega` is connected in series with battery of EMF `3V` and negligible internal resistance and a resistance of `10 Omega`. The potential gradient along the wire is :

Options

A. `1 V//m`
B. `0.5 V//m`
C. `0.2 V//m`
D. `0.1 V//m`

Answer

C

Explanation

`I=(3)/(20+10)=0.1A`
`therefore " Potential gradient " = (dV)/(dx)=(IR)/(L)`
` =(20xx0.1)/(10)=0.2 V//m`