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Step-by-Step Solution
Step 1: Understanding the Setup
A cell is connected to a potentiometer wire. Its voltage is balanced at a certain length of the wire. When an external resistance of 10 $\Omega$ is connected across the cell (i.e., the cell is short-circuited through this resistance), the balancing length changes. We need to find the internal resistance $r$ of the cell.
Step 2: Balancing Without the External Resistance
Let $E$ be the emf of the cell. When the cell is not short-circuited (i.e., no extra resistance), it balances a length $l_1 = 110\text{ cm}$ of the potentiometer wire. If $V$ is the potential drop per cm along the potentiometer wire, then the balance condition implies:
$E = \frac{V}{L} \times l_1
= \frac{V}{L} \times 110.$
Here, $L$ simply denotes the length scale used for the potentiometer’s calibration, and $V/L$ is the potential drop per cm. The exact value of $L$ is not crucial if it remains the same in both measurements.
Step 3: Balancing With the External Resistance
When a resistor $R = 10\,\Omega$ is connected across the cell, the total circuit includes the cell’s internal resistance $r$ and the external resistance $R$ in series when measuring current. The current $i$ through the circuit is:
$i = \frac{E}{R + r}.
This causes the cell’s terminal voltage when delivering current to drop to $E_{\text{terminal}} = E - i\,r.$ However, for the potentiometer balance condition, we use the voltage actually across the cell (which is $E_{\text{terminal}}$). The new balancing length is $l_2 = 100\text{ cm}$. Hence,
$E_{\text{terminal}} = \text{(potential drop per cm)} \times 100
= \frac{V}{L} \times 100.
But $E_{\text{terminal}} = E - i\,r = E - \frac{E}{R + r} \, r = E \left(1 - \frac{r}{R + r}\right) = E \frac{R}{R + r}.$
(Another way to see it is $E_{\text{terminal}} = \frac{R}{R + r}\,E.$)
Therefore:
$\frac{R}{R + r} E = \frac{V}{L} \times 100.
Step 4: Forming the Ratio of the Two Balances
From the first balance (no current load)
$E = \frac{V}{L} \times 110.
From the second balance (with $R$):
$\frac{R}{R + r} E = \frac{V}{L} \times 100.
Divide the first equation by the second to eliminate $E$ and $\frac{V}{L}$:
$ \frac{E}{\frac{R}{R + r} E}
= \frac{\frac{V}{L} \times 110}{\frac{V}{L} \times 100}
\quad \implies \quad
\frac{R + r}{R}
= \frac{110}{100}
= 1.1.
Step 5: Solving for the Internal Resistance
From the equation
$\frac{R + r}{R} = 1.1 \quad \implies \quad R + r = 1.1 \, R \quad \implies \quad r = 1.1\,R - R = 0.1\,R.
Given $R = 10\,\Omega$, we substitute:
$r = 0.1 \times 10\,\Omega = 1\,\Omega.
Final Answer
The internal resistance of the cell is $1\,\Omega.$
