A conductor of mass 0.50 kg and density `9xx10^3kg m^(-3)` has a resistance of `0.015 Omega` . Calculate the length of the conductor if its shape is cylindrical and the resistivity of its material is `1.8 xx10^(-7) Omega m`.

Answer

To solve the problem, we need to find the length of a cylindrical conductor given its mass, density, resistance, and resistivity. Here’s a step-by-step solution:

### Step 1: Calculate the Volume of the Conductor
The volume \( V \) of the conductor can be calculated using the formula:
\[ V = \frac{m}{\rho} \]
where:
- \( m = 0.50 \, \text{kg} \) (mass of the conductor)
- \( \rho = 9 \times 10^3 \, \text{kg/m}^3 \) (density of the conductor)

Substituting the values:
\[ V = \frac{0.50 \, \text{kg}}{9 \times 10^3 \, \text{kg/m}^3} = \frac{0.50}{9000} \approx 5.56 \times 10^{-5} \, \text{m}^3 \]

### Step 2: Relate Volume to Length and Cross-Sectional Area
For a cylindrical conductor, the volume \( V \) can also be expressed in terms of its length \( L \) and cross-sectional area \( A \):
\[ V = A \times L \]

### Step 3: Express Cross-Sectional Area in Terms of Length
From the volume equation, we can express the cross-sectional area \( A \) as:
\[ A = \frac{V}{L} \]

### Step 4: Use the Resistance Formula
The resistance \( R \) of the conductor is given by:
\[ R = \frac{\rho L}{A} \]
Substituting \( A \) from the previous step:
\[ R = \frac{\rho L}{\frac{V}{L}} = \frac{\rho L^2}{V} \]

### Step 5: Rearranging for Length
Rearranging the equation to solve for \( L^2 \):
\[ L^2 = \frac{R \cdot V}{\rho} \]

### Step 6: Substitute Known Values
Now substitute the known values into the equation:
- \( R = 0.015 \, \Omega \)
- \( V = 5.56 \times 10^{-5} \, \text{m}^3 \)
- \( \rho = 1.8 \times 10^{-7} \, \Omega \cdot \text{m} \)

Calculating \( L^2 \):
\[ L^2 = \frac{0.015 \cdot 5.56 \times 10^{-5}}{1.8 \times 10^{-7}} \]
Calculating the numerator:
\[ 0.015 \cdot 5.56 \times 10^{-5} \approx 8.34 \times 10^{-7} \]
Now calculating \( L^2 \):
\[ L^2 = \frac{8.34 \times 10^{-7}}{1.8 \times 10^{-7}} \approx 4.63 \]

### Step 7: Calculate Length
Taking the square root to find \( L \):
\[ L = \sqrt{4.63} \approx 2.15 \, \text{m} \]

### Final Answer
The length of the conductor is approximately \( 2.15 \, \text{m} \).

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