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. 260 `Omega`
B. `760 `Omega``
C. `960`Omega``
D. `1060`Omega``

Answer

B

Explanation

Poteitial gradient alone wire
`=("Poteitial difference along wire")/(Length of wire")`
`therefore 0 1xx10^(-3)=(lxx40)/(1000)V//cm`
Current in wire,
`L=1/400A`
`therefore l=(E)/(R+R')`
`(2)/(40+R)=1/400` ltbr `or R=800-40=760Omega`