Correct Answer - d
Resistance of wire `AB = 0.02 xx 100 = 2 Omega`
Current in wire `AB, I= (2)/(2+2) = (1)/(2)A`
Let `x` be the resistance of length `AJ` of wire of which galvanometer shown on deflection, then
`V_(A)- V_(J)=x I= x xx 1//2`....(i)
Here, `epsilon_(1) = 2V: epsilon_(2) = 1V: r_(1) = 2Omega : r_(2) = 1Omega`
Current closed circiut having `epsilon_(1)` and `epsilon_(2)` in series ,
`I_(1) = (2+1)/(2+1) = 1A`
pot diff across `C` and `D` i.e.
`V_(C)- V_(D) = epsilon_(1) - I_(1)r_(1) = 2- 1 xx 2 = 0`
Here, `V_(A) - V_(J) = V_(C)-V_(D) or (x)/(2) = 0 or x = 0`
i.e. length of wire `AJ= 0`