The length of the potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of emf 2 V and internal resistance `10Omega` is balanced on this potentiometer the balance length is found to be 500 cm. The resistance of potentiometer wire will be

Options

A. `20Omega`
B. `40 Omega`
C. `60Omega`
D. `80Omega`

Answer

To find the resistance of the potentiometer wire, we can follow these steps:

### Step 1: Understand the relationship between EMF, potential gradient, and balance length
The EMF (E) of the cell is related to the potential gradient (V/L) and the balance length (L0) by the formula:
\[ E = \left(\frac{V}{L}\right) \times L_0 \]
Where:
- \( E \) = EMF of the cell (2 V)
- \( V \) = potential difference across the potentiometer wire
- \( L \) = total length of the potentiometer wire (600 cm)
- \( L_0 \) = balance length (500 cm)

### Step 2: Calculate the potential gradient
The potential gradient (V/L) can be expressed in terms of the current (I) flowing through the potentiometer wire and its resistance (R):
\[ V = I \times R \]
Thus, we can rewrite the potential gradient as:
\[ \frac{V}{L} = \frac{I \times R}{L} \]

### Step 3: Substitute the values into the equation
Substituting the expression for V into the EMF equation:
\[ E = \left(\frac{I \times R}{L}\right) \times L_0 \]
Now substituting the known values:
- \( E = 2 \, \text{V} \)
- \( I = 40 \, \text{mA} = 40 \times 10^{-3} \, \text{A} \)
- \( L = 600 \, \text{cm} = 600 \times 10^{-2} \, \text{m} \)
- \( L_0 = 500 \, \text{cm} = 500 \times 10^{-2} \, \text{m} \)

### Step 4: Rearranging the equation to find R
Now we can rearrange the equation to solve for R:
\[ 2 = \left(\frac{40 \times 10^{-3} \times R}{600 \times 10^{-2}}\right) \times (500 \times 10^{-2}) \]

### Step 5: Simplifying the equation
Rearranging gives:
\[ 2 = \frac{40 \times 10^{-3} \times R \times 500 \times 10^{-2}}{600 \times 10^{-2}} \]
\[ 2 = \frac{40 \times 500 \times R}{600} \times 10^{-3} \]
\[ 2 = \frac{20000 \times R}{600} \times 10^{-3} \]
\[ 2 = \frac{20000}{600} \times R \times 10^{-3} \]

### Step 6: Solving for R
Now we can solve for R:
\[ R = \frac{2 \times 600}{20000 \times 10^{-3}} \]
\[ R = \frac{1200}{20} \]
\[ R = 60 \, \Omega \]

Thus, the resistance of the potentiometer wire is **60 ohms**.