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. 3 V/m
B. 0.2 V/m
C. 0.1 V/m
D. 0.3 V/m

Answer

C

Explanation

As potential gradient
`=("Potential difference across wire")/("Length of wire")=(R_("wire")xx l)/(l)` …(i)
Current `l=(E )/(R_("wire")+R_("ext"))=(3)/((20+10))=0.1 V//m`