A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ε and internal resistance r. A cell C having emf ε/2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in fig. shows no deflection is?

Let the potential gradient along the potentiometer wire AB be k.
Then, k = ε/(L+r/R) * R/L = εR/(L(R+r)) where R = 12r
The potential difference across AJ is k * AJ = k * x where x is the length AJ.
The potential difference across the cell C is ε/2. Since there is no deflection, the potential difference across AJ must be equal to the emf of cell C.
Therefore, kx = ε/2
Substituting the value of k, we have:
(εR/(L(R+r))) * x = ε/2
(12rε)/(L(12r+r)) * x = ε/2
(12r)/(13L) * x = 1/2
x = (13L)/(24)
x = 13/24L
Therefore, the length AJ at which the galvanometer shows no deflection is 13/24L.