The total length of potentiometer wire AB is 50 cm in the arrangement as shown in figure. If $P$ is the point where the galvanometer shows zero reading then the length $A P$ is $\_\_\_\_$ cm.

In this potentiometer, for zero galvanometer reading, the potential at P must equal the potential at the junction between the 6Ω and 4Ω resistors.
The voltage distribution across the resistors is $V_{junction} = E \times \frac{6}{6+4} = 0.6E$ from point A.
Since the potentiometer wire is uniform with length 50 cm, the potential distribution is also uniform.
Therefore, $\frac{AP}{AB} = \frac{6}{10}$, giving $AP = 50 \times 0.6 = 30$ cm.