The specific resistance of a potentiometer wire is `10^(-7) Omega`m and the cross sectional area of the wire is `10^(-6)m^(2)`. If a current of 0.1 A flows through the wire, then the potential gradient will be

Options

A. `10^(-4) V//m`
B. `10^(-3)V//m`
C. `10^(-2) V//m`
D. `10^(-5)V//m`

Answer

C

Explanation

`R=(el)/(A)=(10^(-7)xxl)/(10^(-6))=0.1 l therefore (R )/(l)=0.1`
`therefore (V)/(l)=(IR)/(l)=0.1xx0.1 =10^(-2) V//m`