A potentiometer wire 10 long has a resistance of `40Omega`. It is connected in series with a resistances box and a 2 v storage cell. If the potential gradient along the wire is `0.01(V)/(m)` the resistance unplugged in the box is

Options

A. `560 Omega`
B. `760 Omega`
C. `960 Omega`
D. `660 Omega`

Answer

B

Explanation

Given : Potential gradient,
`K= (0.1 mv)/(cm)=(0.1xx10^(-3)V)/(10^(-2)m)`
`= 10^(-2) V//m =(1)/(100) V//m`
and `I= (E )/(R+40)=(2)/(R+40)`
and L= 10 m and `R' = 40 Omega`
`therefore K=(V)/(L)=(IR')/(L)=((2)/(R+40))(40)/(10)=(8)/(R+40)`
`therefore (1)/(100)=(8)/(R+40)`
`therefore 800 = R+40 therefore R= 800-40=760 Omega`