When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found to be 540 cm. If the balancing length becomes 500 cm. When the cell is short-circuited with `1Omega`, the internal resistance of the cell is

Answer

To find the internal resistance of the Daniel cell when it is short-circuited, we can follow these steps:

### Step 1: Understand the initial conditions
When the Daniel cell is connected to the potentiometer, the balancing length is 540 cm. This means that the potential difference (E) across the cell can be expressed as:
\[ E = V_0 \times L_1 \]
where \( V_0 \) is the potential drop per unit length of the potentiometer wire and \( L_1 = 540 \, \text{cm} \).

### Step 2: Understand the conditions when the cell is short-circuited
When the cell is short-circuited with a 1 ohm resistor, the balancing length changes to 500 cm. The potential difference across the battery can now be expressed as:
\[ E = V_0 \times L_2 \]
where \( L_2 = 500 \, \text{cm} \).

### Step 3: Set up the equations
From the two conditions, we can set up the following equations:
1. \( E = V_0 \times 540 \)
2. \( E = I \times (R + 1) \) where \( I \) is the current through the circuit and \( R \) is the internal resistance of the cell.

### Step 4: Calculate the current when short-circuited
The current \( I \) can be expressed as:
\[ I = \frac{E}{R + 1} \]

### Step 5: Relate the two equations
From the second condition, we can express the potential difference across the external circuit (1 ohm resistor) as:
\[ E = I \times 1 = I \]

Substituting \( I \) from the previous equation gives:
\[ E = \frac{E}{R + 1} \]

### Step 6: Substitute the values of balancing lengths
Now, we can relate the two expressions for \( E \):
\[ \frac{V_0 \times 540}{V_0 \times 500} = \frac{1}{R + 1} \]

This simplifies to:
\[ \frac{540}{500} = \frac{1}{R + 1} \]

### Step 7: Solve for \( R \)
Cross-multiplying gives:
\[ 540(R + 1) = 500 \]
\[ 540R + 540 = 500 \]
\[ 540R = 500 - 540 \]
\[ 540R = -40 \]
\[ R = \frac{-40}{540} = -\frac{2}{27} \]

### Step 8: Correct the approach
Since we are looking for internal resistance, we should have:
\[ R + 1 = \frac{540}{500} \]
\[ R + 1 = \frac{27}{25} \]
\[ R = \frac{27}{25} - 1 = \frac{27 - 25}{25} = \frac{2}{25} \]

### Step 9: Final answer
Thus, the internal resistance \( R \) of the cell is:
\[ R = 0.08 \, \Omega \]