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. 0.02
B. 0.1
C. 0.2
D. 1.2

Answer

C

Explanation

As resistances are in series,
`R_("total")=R_(1)+R_(2)=20+10=30 Omega`
`i=(4)/(R_("total"))=(3)/(30)=(1)/(10)A`
So, `V_("wire")=iR_("wire")=(1)/(10)xx20=2V`
Hence, potential gradient is `(V_("wire"))/(l)=(2)/(10)=0.2 Vm^(-1)`