The total length of potentiometer wire 𝐴𝐵 is 50 cm in the arrangement so in the in fig. If ' 𝑃′ is the point where the galvanometer shows zero reading Then the length AP is ____ cm .

The correct answer is 30.

Given Data:

• Total length of potentiometer wire AB, $L=50\text{ cm}$
• Resistance of the upper left arm, $R_1=6\Omega$
• Resistance of the upper right arm, $R_2=4\Omega$

Understanding the Circuit:

The given setup forms a Wheatstone bridge circuit balanced by the potentiometer wire AB. When the galvanometer shows zero deflection (null point) at point P, the potential difference across the resistance $R_1$ is equal to the potential difference across the segment AP of the wire.

Since the wire AB is uniform, its resistance is directly proportional to its length ($R\propto l$).

Condition for Null Deflection:

Using the principle of a balanced Wheatstone bridge:

$\frac{R_1}{R_2}=\frac{l_{AP}}{l_{PB}}$

Let $l_{AP}=x\text{ cm}$. Then the remaining length of the wire is:

$l_{PB}=(50-x)\text{ cm}$

Step-by-Step Calculation:

Substitute the given values into the balancing condition:

$\frac{6}{4}=\frac{x}{50-x}$

Cross-multiplying to solve for $x$:

$6\times (50-x)=4x$

$300-6x=4x$

$10x=300$

$x=30\text{ cm}$

Conclusion:

The length of segment AP where the galvanometer shows zero reading is 30 cm.