To find the resistance of the potentiometer wire, we can follow these steps:
### Step 1: Understand the given information
- Length of the potentiometer wire (L) = 1200 cm
- Current (I) = 60 mA = 0.060 A
- EMF of the cell (E) = 5 V
- Internal resistance of the cell (r) = 20 Ω
- Null point distance (d) = 1000 cm
### Step 2: Determine the potential difference across the null point
Since the null point is at 1000 cm, we can find the potential difference across this length of the wire. The potential difference across the entire length of the wire can be calculated using the ratio of the lengths.
The potential difference (V) across the null point can be calculated as follows:
\[ V = \frac{d}{L} \times E \]
\[ V = \frac{1000 \, \text{cm}}{1200 \, \text{cm}} \times 5 \, \text{V} \]
\[ V = \frac{1000}{1200} \times 5 \]
\[ V = \frac{5}{1.2} \]
\[ V = 4.1667 \, \text{V} \]
### Step 3: Calculate the total resistance in the circuit
Using Ohm's Law, the total resistance (R_total) in the circuit can be calculated as:
\[ R_{\text{total}} = \frac{E}{I} \]
Where E is the total EMF and I is the current flowing through the circuit.
The total EMF is the sum of the potential difference across the potentiometer wire and the voltage drop across the internal resistance of the cell:
\[ E = V + I \cdot r \]
\[ E = 4.1667 \, \text{V} + (0.060 \, \text{A} \cdot 20 \, \Omega) \]
\[ E = 4.1667 \, \text{V} + 1.2 \, \text{V} \]
\[ E = 5.3667 \, \text{V} \]
Now, we can calculate the total resistance:
\[ R_{\text{total}} = \frac{5.3667 \, \text{V}}{0.060 \, \text{A}} \]
\[ R_{\text{total}} = 89.445 \, \Omega \]
### Step 4: Calculate the resistance of the potentiometer wire
The resistance of the potentiometer wire (R_wire) can be found by subtracting the internal resistance of the cell from the total resistance:
\[ R_{\text{wire}} = R_{\text{total}} - r \]
\[ R_{\text{wire}} = 89.445 \, \Omega - 20 \, \Omega \]
\[ R_{\text{wire}} = 69.445 \, \Omega \]
### Conclusion
The resistance of the potentiometer wire is approximately 69.45 Ω.
