The resistance of the meter bridge AB in given figure is 4 `Omega` .With a cell of emf `epsilon=0.5 V` and rheostat resistance `R_(h) = 2 Omega ` the null point is obatined at some point J is found for `R_(h) = 6 Omega " The emf "epsilon _(2)` is

Options

A. 0.3V
B. 0.4V
C. 0.5V
D. 0.6V

Explanation

At pull point Current `i=6/((R_h+4))`
For balance e=I
Where x is the resistance between A and I
SO case I `0.5(6x)/(2+4) implies x=0.5`
Case 2 `e_2=(6x)/10 implies e_2=0.3` volts