A battery of 6 volts is connected ot the termainals of a three meter long wire of uniform thickness and resistance of the order of `100 Omega`. The difference of potential between two points separated by `50 cm` on the wire will

Options

A. `2 V`
B. `3 V`
C. `1 V`
D. `1.5 V`

Answer

C

Explanation

(c ) Total current drawn from the battery
`i= (E)/(R + r) = (6)/(100 + 0) = 0.06 A`
Resistance of `50 cm` wire is `R' = (rho l')/(A) = ((rho)/(A)) l'`
`= ((R )/(l)) l'` (`:' R = (rho l)/(A))`
`= (100)/(300) xx 50`
So, `R' = (50)/(30) Omega`
Hence, the potential between two point on the wire spareted by a distance `l` is
`V = iR = 0.06 xx (50)/(3) = 1 V`