A 6 V battery is connected to the terminals of a 3 m long wire of uniform thickness and resistance of `100 Omega`. The difference of potential between two points on the wire separated by a distance of 50 cm will be

Answer

`sqrt(3gx)/(8)`

Explanation

`2 mg - T = 2ma, T - mg = ma, a = (g)/(3)`

Velocity of m & `2m` after falling through a distance `x = sqrt(2ax) = sqrt((2gx)/(3))`
Impulse equation
`T Deltat = 2m (v - sqrt((2gx)/(3)))`
`T Deltat - T' Deltat = m (v - sqrt((2gx)/(3)))`
`T' Deltat = m(v- 0), v = sqrt((3gx)/(8))`