A cell can be balanced against `110 cm` and `100 cm` of potentiometer wire, respectively with and without being short circuited through a resistance of `10 Omega`. Its internal resistance is

Options

A. `1.0Omega`
B. `0.5Omega`
C. `2.0Omega`
D. zero

Answer

To find the internal resistance of the cell, we can follow these steps:

### Step 1: Understand the problem
We have a cell that can be balanced against two different lengths of a potentiometer wire: 110 cm when the cell is not short-circuited, and 100 cm when it is short-circuited through a resistance of 10 Ω.

### Step 2: Set up the relationship
When the cell is connected to the potentiometer wire, the electromotive force (E) of the cell can be expressed in terms of the lengths of the wire (L1 and L2) and the resistances involved. The relationship can be given as:
\[ E = k \cdot L \]
where \( k \) is a constant of proportionality and \( L \) is the length of the wire.

### Step 3: Define the lengths
Let:
- \( L_1 = 110 \) cm (length when not short-circuited)
- \( L_2 = 100 \) cm (length when short-circuited)

### Step 4: Set up the equations
When the cell is not short-circuited:
\[ E = k \cdot 110 \]
When the cell is short-circuited through a resistance \( R = 10 \, \Omega \):
\[ E - I \cdot r = k \cdot 100 \]
where \( r \) is the internal resistance of the cell and \( I \) is the current flowing through the circuit.

### Step 5: Relate the two scenarios
From the first scenario:
\[ E = k \cdot 110 \]
From the second scenario:
\[ E = k \cdot 100 + I \cdot r \]

### Step 6: Express current in terms of resistance
Using Ohm's law, the current \( I \) can be expressed as:
\[ I = \frac{E}{R + r} \]
Substituting this into the second equation gives:
\[ E = k \cdot 100 + \frac{E}{R + r} \cdot r \]

### Step 7: Substitute known values
Now, we can substitute \( R = 10 \, \Omega \) into the equations. We also know that:
\[ \frac{L_1 - L_2}{L_2} = \frac{r}{R} \]
Thus:
\[ \frac{110 - 100}{100} = \frac{r}{10} \]
This simplifies to:
\[ \frac{10}{100} = \frac{r}{10} \]
\[ \frac{1}{10} = \frac{r}{10} \]

### Step 8: Solve for internal resistance \( r \)
Cross-multiplying gives:
\[ r = 1 \, \Omega \]

### Final Result
The internal resistance of the cell is \( r = 1 \, \Omega \).

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