Animated Solution for Physics - Current Electricity: In a meter bridge experiment, null point is obtained at $20\text{ cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X<Y$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of $4X$ against $Y$?

Imagine you are standing in a laboratory, looking at a simple yet incredibly elegant piece of apparatus: the meter bridge. It consists of a uniform wire exactly one meter (100 cm) long, stretched taut across a wooden board. This device is a practical application of the famous Wheatstone bridge principle, allowing us to determine unknown resistances with remarkable precision.

The core principle is beautifully simple. When the galvanometer shows zero deflection (the null point), the bridge is balanced. At this magical point, the ratio of the resistances in the left and right gaps is perfectly equal to the ratio of the lengths of the wire segments on either side of the jockey.

Mathematically, this is expressed as:

where $l$ is the balancing length from the left end, and $100-l$ is the remaining length of the wire.

Let's dive into the first part of our problem. We are given that a resistance $X$ is placed in the left gap and a resistance $Y$ is placed in the right gap. The null point is found at a distance of $20\text{ cm}$ from the left end.

By plugging these values into our master equation, we get:

This tells us a crucial piece of information: the resistance $Y$ is exactly four times the resistance $X$. Keep this ratio safe; it is the key to unlocking the rest of the problem.

Now, the experiment takes a twist. We decide to replace the resistance $X$ in the left gap with a new resistance that is four times larger, i.e., $4X$. The resistance $Y$ in the right gap remains untouched.

Because we have changed the resistance on one side, the balance of the bridge is disturbed. The null point must shift to a new position to restore equilibrium. Let's call this new balancing length $l^{\prime }$.

Our new balancing equation becomes:

This equation might look a bit intimidating, but remember that we already know the value of the ratio $\frac{X}{Y}$. Let's substitute the value we found earlier ($\frac{1}{4}$) into our new equation:

The $4$ and the $\frac{1}{4}$ cancel each other out perfectly, leaving us with a beautifully simple equation:

Now, we just cross-multiply to solve for $l^{\prime }$:

Think about the physical intuition here: by increasing the left resistance to $4X$, we made it exactly equal to the right resistance $Y$ (since $Y=4X$). When the resistances in both gaps are identical, the bridge must balance exactly in the middle of the wire. The math perfectly aligns with the physical reality!