The total length of potentiometer wire AB is 50 cm in the arrangement as shown in figure. If P is the point where the ga

Question

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 AP is cm.

Accepted Answer

Given: The total length of potentiometer wire AB is $50$ cm. The resistors are $6 \, \Omega$ and $4 \, \Omega$ . The galvanometer shows zero reading at point P.

Find: The length AP.

At null deflection, the bridge is balanced. Therefore, the ratio of resistances equals the ratio of the lengths of the potentiometer wire segments.

\frac{R_1}{R_2} = \frac{L_{AP}}{L_{PB}}

Let $L_{AP} = x$ cm. Then

L_{PB} = 50 - x

Using $R_1 = 6 \, \Omega$ and $R_2 = 4 \, \Omega$ , we get

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

Cross-multiplying,

6(50 - x) = 4x

300 - 6x = 4x

10x = 300

x = 30

Therefore, the length AP is $30$ cm. The correct option is A.

Given: The galvanometer shows zero reading at P.

Find: Why the wire length can be related to the resistor ratio.

When the galvanometer shows zero reading, the potential at point P is equal to the potential at the junction of the $6 \, \Omega$ and $4 \, \Omega$ resistors. This is the balanced Wheatstone bridge condition.

Since the potentiometer wire is uniform, resistance is directly proportional to length. Hence,

\frac{R_{AP}}{R_{PB}} = \frac{L_{AP}}{L_{PB}}

So the resistor ratio on the top branch must match the wire-segment ratio on the lower branch.

Thus,

\frac{6}{4} = \frac{L_{AP}}{L_{PB}}

with

L_{AP} + L_{PB} = 50

Solving gives

L_{AP} = 30 \text{ cm}

Hence, the correct option is A.

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