There is a potentiometer wire of length 1200 cm and 60 mA current is flowing in it. A battery of emf 5V and internal resistance of `20 ohm` is balanced on potentiometer wire with balancing length 1000 cm. The resistance of potentiometer wire is

Options

A. `60Omega`
B. `100Omega`
C. `120Omega`
D. `80Omega`

Answer

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

### Step 1: Understand the given data
- Length of the potentiometer wire, \( L = 1200 \, \text{cm} \)
- Current flowing through the wire, \( I = 60 \, \text{mA} = 0.06 \, \text{A} \)
- EMF of the battery, \( E = 5 \, \text{V} \)
- Internal resistance of the battery, \( r = 20 \, \Omega \)
- Balancing length on the potentiometer wire, \( l = 1000 \, \text{cm} \)

### Step 2: Calculate the potential drop across the balancing length
The potential drop across the balancing length of the potentiometer wire can be expressed as:
\[ V = E = 5 \, \text{V} \]
This potential drop corresponds to a length of \( 1000 \, \text{cm} \).

### Step 3: Find the potential drop per unit length
The potential drop per unit length (\( V_{l} \)) can be calculated as:
\[ V_{l} = \frac{V}{l} = \frac{5 \, \text{V}}{1000 \, \text{cm}} = 0.005 \, \text{V/cm} \]

### Step 4: Calculate the total potential drop across the entire wire
Now, we can find the total potential drop across the entire length of the potentiometer wire (1200 cm):
\[ V_{total} = V_{l} \times L = 0.005 \, \text{V/cm} \times 1200 \, \text{cm} = 6 \, \text{V} \]

### Step 5: Use Ohm's Law to find the resistance of the potentiometer wire
Using Ohm's Law, the resistance \( R \) of the potentiometer wire can be calculated as:
\[ R = \frac{V_{total}}{I} = \frac{6 \, \text{V}}{0.06 \, \text{A}} = 100 \, \Omega \]

### Final Answer
The resistance of the potentiometer wire is \( R = 100 \, \Omega \).

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