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

`V=IR rArr I=(V)/(R )`
`=(3)/((20+10)) =(3)/(30)=(1)/(10)A`
Potential gradient `=((DeltaV)/(Deltal))`
Potential across `20Omega` resistance,
`V=lR =(1)/(10) xx 20=2V`
`:. DeltaV//Deltal=2//10=0.2 V//m`