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

Answer

C

Explanation

At pull point , current `i=6/((R_h+4))`
For balance : `epsilon=i`
Where x is the resistance between A and I.
So, Case 1 : `0.5=(6x)/(2+4) rArr ` x=0.5
Case 2: `epsilon_2=(6x)/10 rArr epsilon_2` =0.3 volts .