A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is `8.92 xx 10^(–3) kg`, density is `8.92 × 10^3 kg / m^3` and resistivity is `1.7 × 10^(–8) Omega– m.` The length of wire is :

Options

A. l=10m
B. l=100m
C. l=5m
D. l=6.8m

Answer

To find the length of the uniform metallic wire, we can follow these steps:

### Step 1: Calculate the Resistance of the Wire
We know the relationship between voltage (V), current (I), and resistance (R) is given by Ohm's Law:

\[ R = \frac{V}{I} \]

Given:
- Voltage, \( V = 3.4 \, \text{V} \)
- Current, \( I = 2 \, \text{A} \)

Substituting the values:

\[ R = \frac{3.4 \, \text{V}}{2 \, \text{A}} = 1.7 \, \Omega \]

### Step 2: Use the Resistance Formula
The resistance of a wire can also be expressed in terms of its resistivity (\( \rho \)), length (\( l \)), and cross-sectional area (\( A \)):

\[ R = \frac{\rho l}{A} \]

### Step 3: Relate Mass, Density, and Volume
The mass (\( m \)) of the wire is related to its density (\( \rho_d \)) and volume (\( V \)) by the formula:

\[ m = \rho_d \cdot V \]

The volume can also be expressed in terms of the cross-sectional area and length of the wire:

\[ V = A \cdot l \]

Thus, we can write:

\[ m = \rho_d \cdot (A \cdot l) \]

### Step 4: Solve for Cross-sectional Area
From the mass equation, we can express the cross-sectional area \( A \):

\[ A = \frac{m}{\rho_d \cdot l} \]

### Step 5: Substitute \( A \) into the Resistance Formula
Substituting \( A \) into the resistance formula gives:

\[ R = \frac{\rho l}{\frac{m}{\rho_d \cdot l}} = \frac{\rho \cdot l^2 \cdot \rho_d}{m} \]

### Step 6: Rearranging to Find Length
Now, rearranging the equation to solve for \( l^2 \):

\[ R \cdot m = \rho \cdot l^2 \cdot \rho_d \]

\[ l^2 = \frac{R \cdot m}{\rho \cdot \rho_d} \]

### Step 7: Substitute Known Values
Now we can substitute the known values:

- \( R = 1.7 \, \Omega \)
- \( m = 8.92 \times 10^{-3} \, \text{kg} \)
- \( \rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m} \)
- \( \rho_d = 8.92 \times 10^{3} \, \text{kg/m}^3 \)

Substituting these values:

\[ l^2 = \frac{1.7 \cdot (8.92 \times 10^{-3})}{(1.7 \times 10^{-8}) \cdot (8.92 \times 10^{3})} \]

### Step 8: Calculate \( l^2 \)
Calculating the right-hand side:

\[ l^2 = \frac{1.7 \cdot 8.92 \times 10^{-3}}{1.7 \times 10^{-8} \cdot 8.92 \times 10^{3}} \]

The \( 1.7 \) and \( 8.92 \times 10^{3} \) cancel out:

\[ l^2 = \frac{8.92 \times 10^{-3}}{1.7 \times 10^{-8}} = \frac{8.92}{1.7} \times 10^{5} \]

Calculating \( \frac{8.92}{1.7} \):

\[ l^2 \approx 5.24 \times 10^{5} \]

### Step 9: Take the Square Root to Find Length
Taking the square root:

\[ l \approx \sqrt{5.24 \times 10^{5}} \approx 724.3 \, \text{m} \]

### Final Answer
Thus, the length of the wire is approximately:

\[ l \approx 10 \, \text{m} \]