Animated Solution for Physics - Current Electricity: The balancing length for a cell is $560\text{ cm}$ in a potentiometer experiment. When an external resistance of $10\Omega$ is connected in parallel to the cell, the balancing length changes by $60\text{ cm}$. If the internal resistance of the cell is $\frac{N}{10}\Omega$, where $N$ is an integer, then value of $N$ is ............ .

Imagine you are trying to measure the exact weight of a delicate feather, but your weighing scale is so heavy that just placing the feather on it changes the reading. This is exactly what happens when you use a standard voltmeter to measure the Electromotive Force (EMF) of a cell. The voltmeter draws a tiny amount of current, which causes a voltage drop across the cell's internal resistance, giving you a slightly lower reading known as the terminal voltage.

Enter the potentiometer—a beautifully elegant instrument that measures EMF by drawing absolutely zero current from the test cell at the balance point. It works on the principle of opposing potentials. When the potential drop across a specific length of the potentiometer wire exactly matches the cell's EMF, no current flows through the galvanometer.

In our specific problem, we are using this magical instrument to find the internal resistance $r$ of a cell. The problem gives us two distinct scenarios:

1. The Open Circuit (Measuring EMF): Initially, the cell is connected alone. The galvanometer finds a null point at a balancing length $l_1=560\text{ cm}$. At this point, the potential drop across $560\text{ cm}$ of the wire is exactly equal to the cell's true EMF, $E$. So, we can write $E\propto l_1$.

2. The Closed Circuit (Measuring Terminal Voltage): Next, a known external resistance $R=10\Omega$ is connected in parallel to the cell. Now, the cell starts supplying current to this external resistor. Because current is flowing, the potential difference across the cell drops from its full EMF $E$ to its terminal voltage $V$.

The problem states that the balancing length changes by $60\text{ cm}$. Here is where many students make a silly mistake! Does it increase or decrease? Since the terminal voltage $V$ is always less than the EMF $E$ (because $V=E-ir$), the new balancing length must be shorter.

Therefore, the new balancing length is $l_2=560-60=500\text{ cm}$. This length corresponds to the terminal voltage, so $V\propto l_2$.

To find the internal resistance, we use the standard relationship between EMF, terminal voltage, external resistance, and internal resistance:

Since $E$ and $V$ are directly proportional to their respective balancing lengths $l_1$ and $l_2$, we can substitute them directly into the formula:

Now, it is just a matter of careful substitution and basic algebra. Let's plug in our values: $R=10\Omega$, $l_1=560\text{ cm}$, and $l_2=500\text{ cm}$.

Simplifying the fraction inside the bracket:

Taking the common denominator:

The $10$ and the $50$ cancel out beautifully, leaving us with:

The problem asks us to express this internal resistance in the form of $\frac{N}{10}\Omega$. So, we simply equate our result to this expression:

Cross-multiplying to solve for $N$:

And there we have it! The integer value of $N$ is 12. By understanding the physical reality behind the balancing lengths, the math flows naturally without any confusion.