Animated Solution for Physics - Current Electricity: In the shown arrangement of the experiment of the meter bridge if $AC$ corresponding to null deflection of galvanometer is $x$, what would be its value if the radius of the wire $AB$ is doubled ?

Imagine you are in a physics laboratory, standing in front of a classic meter bridge setup. It’s a beautifully simple yet incredibly precise instrument used to measure unknown resistances. At its heart lies a long, uniform wire, typically one meter in length, stretched taut between two points, $A$ and $B$.

When we connect a galvanometer and slide the jockey along this wire, we are essentially hunting for a point of perfect electrical symmetry. This magical spot is called the null point, denoted by $C$. At this exact location, the potential difference across the galvanometer drops to absolute zero, and the needle stands perfectly still.

Why does the galvanometer show zero deflection? It all comes down to the principle of the balanced Wheatstone bridge. When the bridge is balanced, the ratio of the resistances in the upper arms of the circuit is exactly equal to the ratio of the resistances of the two segments of our wire.

Mathematically, we can express this elegant relationship as:

Here, $R_{AC}$ is the resistance of the wire segment from $A$ to $C$, and $R_{CB}$ is the resistance from $C$ to $B$. The beauty of this equation is that it links the abstract concept of resistance directly to the physical lengths of the wire segments.

To understand how the physical dimensions of the wire affect our null point, we need to dive into the formula for the resistance of a uniform conductor. The resistance $R$ of any wire is directly proportional to its length $l$ and inversely proportional to its cross-sectional area $A$.

This is given by the formula:

Where $\rho$ is the resistivity of the material. For our meter bridge, the length of segment $AC$ is $x$, and the length of segment $CB$ is $L-x$. Therefore, we can write their respective resistances as:

Now, let's substitute these expressions back into our balance condition. This is where the magic happens!

Look closely at this equation. The resistivity $\rho$ and the cross-sectional area $A$ appear in both the numerator and the denominator. Because the wire is uniform, these values are identical for both segments.

Therefore, they cancel out completely, leaving us with a beautifully simplified equation:

This final equation reveals a profound truth about the meter bridge: the balance condition depends only on the lengths of the wire segments. It is completely independent of the wire's cross-sectional area or its resistivity!

So, what happens if we double the radius of the wire $AB$? Doubling the radius would increase the cross-sectional area by a factor of four ($A=\pi r^2$). However, as we just proved, the area $A$ cancels out of our equation entirely.

The ratio $\frac{x}{L-x}$ remains perfectly unchanged. Consequently, the null point $x$ stays exactly where it was. The correct answer is (a) $x$.

This is why the meter bridge is such a robust and reliable instrument—as long as the wire is uniform, its thickness doesn't matter at all!