How to solve resistivity in numericals class 10?
2 Answers
To solve problems involving resistivity in Class 10 physics, you'll need to understand the concept of resistivity, the relevant formulas, and how to apply them in various numerical scenarios. Here’s a detailed breakdown to help you grasp these concepts effectively:
### Key Concepts
1. **Resistivity (\( \rho \))**:
- It is a property of the material that quantifies how strongly it resists the flow of electric current.
- The unit of resistivity is ohm-meter (\( \Omega \cdot m \)).
2. **Resistance (\( R \))**:
- Resistance is the opposition to the flow of current in a conductor.
- The formula to calculate resistance is:
\[
R = \frac{V}{I}
\]
- Where \( V \) is the voltage (in volts) and \( I \) is the current (in amperes).
3. **Length and Cross-Sectional Area**:
- The resistance of a uniform material depends on its length and cross-sectional area:
\[
R = \rho \frac{L}{A}
\]
- Where:
- \( R \) = resistance (in ohms, \( \Omega \))
- \( \rho \) = resistivity (in ohm-meter, \( \Omega \cdot m \))
- \( L \) = length of the conductor (in meters)
- \( A \) = cross-sectional area of the conductor (in square meters)
### Steps to Solve Numericals Involving Resistivity
1. **Identify Given Values**:
- Read the problem carefully to find values for resistance (\( R \)), length (\( L \)), and area (\( A \)). You may also need to convert units where necessary.
2. **Choose the Appropriate Formula**:
- If you are asked to find resistivity, rearrange the formula:
\[
\rho = R \frac{A}{L}
\]
- If you are given resistivity and asked to find resistance, you would use:
\[
R = \rho \frac{L}{A}
\]
3. **Substitute Values**:
- Plug in the known values into the formula.
4. **Calculate**:
- Perform the calculations carefully, ensuring to maintain the correct units.
5. **Check Your Answer**:
- Ensure that your answer makes sense in the context of the problem.
### Example Problems
Here are a couple of example problems to illustrate:
#### Example 1: Finding Resistivity
**Problem**: A copper wire has a resistance of \( 5 \, \Omega \), a length of \( 2 \, m \), and a cross-sectional area of \( 1 \times 10^{-6} \, m^2 \). Find the resistivity of copper.
**Solution**:
1. Given:
- \( R = 5 \, \Omega \)
- \( L = 2 \, m \)
- \( A = 1 \times 10^{-6} \, m^2 \)
2. Use the formula for resistivity:
\[
\rho = R \frac{A}{L}
\]
3. Substitute the values:
\[
\rho = 5 \times \frac{1 \times 10^{-6}}{2} = 2.5 \times 10^{-6} \, \Omega \cdot m
\]
#### Example 2: Finding Resistance
**Problem**: A wire has a resistivity of \( 1.7 \times 10^{-8} \, \Omega \cdot m \), a length of \( 3 \, m \), and a cross-sectional area of \( 2 \times 10^{-6} \, m^2 \). Calculate its resistance.
**Solution**:
1. Given:
- \( \rho = 1.7 \times 10^{-8} \, \Omega \cdot m \)
- \( L = 3 \, m \)
- \( A = 2 \times 10^{-6} \, m^2 \)
2. Use the formula for resistance:
\[
R = \rho \frac{L}{A}
\]
3. Substitute the values:
\[
R = 1.7 \times 10^{-8} \times \frac{3}{2 \times 10^{-6}} = 0.0255 \, \Omega
\]
### Conclusion
By understanding the definitions and relationships between resistivity, resistance, length, and area, you can tackle a variety of numerical problems in your Class 10 physics curriculum. Practice different types of problems to build confidence and proficiency in solving resistivity-related questions!
### Key Concepts
1. **Resistivity (\( \rho \))**:
- It is a property of the material that quantifies how strongly it resists the flow of electric current.
- The unit of resistivity is ohm-meter (\( \Omega \cdot m \)).
2. **Resistance (\( R \))**:
- Resistance is the opposition to the flow of current in a conductor.
- The formula to calculate resistance is:
\[
R = \frac{V}{I}
\]
- Where \( V \) is the voltage (in volts) and \( I \) is the current (in amperes).
3. **Length and Cross-Sectional Area**:
- The resistance of a uniform material depends on its length and cross-sectional area:
\[
R = \rho \frac{L}{A}
\]
- Where:
- \( R \) = resistance (in ohms, \( \Omega \))
- \( \rho \) = resistivity (in ohm-meter, \( \Omega \cdot m \))
- \( L \) = length of the conductor (in meters)
- \( A \) = cross-sectional area of the conductor (in square meters)
### Steps to Solve Numericals Involving Resistivity
1. **Identify Given Values**:
- Read the problem carefully to find values for resistance (\( R \)), length (\( L \)), and area (\( A \)). You may also need to convert units where necessary.
2. **Choose the Appropriate Formula**:
- If you are asked to find resistivity, rearrange the formula:
\[
\rho = R \frac{A}{L}
\]
- If you are given resistivity and asked to find resistance, you would use:
\[
R = \rho \frac{L}{A}
\]
3. **Substitute Values**:
- Plug in the known values into the formula.
4. **Calculate**:
- Perform the calculations carefully, ensuring to maintain the correct units.
5. **Check Your Answer**:
- Ensure that your answer makes sense in the context of the problem.
### Example Problems
Here are a couple of example problems to illustrate:
#### Example 1: Finding Resistivity
**Problem**: A copper wire has a resistance of \( 5 \, \Omega \), a length of \( 2 \, m \), and a cross-sectional area of \( 1 \times 10^{-6} \, m^2 \). Find the resistivity of copper.
**Solution**:
1. Given:
- \( R = 5 \, \Omega \)
- \( L = 2 \, m \)
- \( A = 1 \times 10^{-6} \, m^2 \)
2. Use the formula for resistivity:
\[
\rho = R \frac{A}{L}
\]
3. Substitute the values:
\[
\rho = 5 \times \frac{1 \times 10^{-6}}{2} = 2.5 \times 10^{-6} \, \Omega \cdot m
\]
#### Example 2: Finding Resistance
**Problem**: A wire has a resistivity of \( 1.7 \times 10^{-8} \, \Omega \cdot m \), a length of \( 3 \, m \), and a cross-sectional area of \( 2 \times 10^{-6} \, m^2 \). Calculate its resistance.
**Solution**:
1. Given:
- \( \rho = 1.7 \times 10^{-8} \, \Omega \cdot m \)
- \( L = 3 \, m \)
- \( A = 2 \times 10^{-6} \, m^2 \)
2. Use the formula for resistance:
\[
R = \rho \frac{L}{A}
\]
3. Substitute the values:
\[
R = 1.7 \times 10^{-8} \times \frac{3}{2 \times 10^{-6}} = 0.0255 \, \Omega
\]
### Conclusion
By understanding the definitions and relationships between resistivity, resistance, length, and area, you can tackle a variety of numerical problems in your Class 10 physics curriculum. Practice different types of problems to build confidence and proficiency in solving resistivity-related questions!
To solve numericals related to resistivity for Class 10, you need to understand the key concepts and formulas associated with it. Here’s a step-by-step guide to help you tackle these problems:
### Key Concepts
1. **Resistivity (\(\rho\))**: Resistivity is a property of a material that indicates how much it resists the flow of electric current. It depends on the material and is defined by the equation:
\[
\rho = R \times \frac{A}{L}
\]
where \( R \) is the resistance, \( A \) is the cross-sectional area of the conductor, and \( L \) is the length of the conductor.
2. **Resistance (R)**: The resistance of a conductor is given by:
\[
R = \rho \times \frac{L}{A}
\]
Rearranging this equation gives us the resistivity:
\[
\rho = R \times \frac{A}{L}
\]
### Steps to Solve Numericals
1. **Understand the Problem**: Read the problem carefully to identify what is given and what needs to be found. Typically, you might be given values for resistance, length, and cross-sectional area and asked to find the resistivity, or vice versa.
2. **Write Down the Formula**: Based on what you need to find, write down the relevant formula. For resistivity, use:
\[
\rho = R \times \frac{A}{L}
\]
3. **Substitute the Values**: Plug in the given values into the formula. Make sure that the units are consistent. For example, if the length is given in meters, the area should be in square meters.
4. **Calculate**: Perform the arithmetic operations to solve for resistivity. Ensure that you carry out the calculations carefully to avoid mistakes.
5. **Check Units**: Ensure that the result has the correct units for resistivity, which is ohm-meters (\(\Omega \cdot m\)).
### Example Problem
**Problem**: A copper wire has a resistance of 2 ohms, a length of 1.5 meters, and a cross-sectional area of \(1 \times 10^{-6}\) square meters. Find the resistivity of copper.
**Solution**:
1. **Identify the Given Values**:
- Resistance \( R = 2 \, \Omega \)
- Length \( L = 1.5 \, \text{m} \)
- Cross-sectional Area \( A = 1 \times 10^{-6} \, \text{m}^2 \)
2. **Use the Resistivity Formula**:
\[
\rho = R \times \frac{A}{L}
\]
3. **Substitute the Values**:
\[
\rho = 2 \, \Omega \times \frac{1 \times 10^{-6} \, \text{m}^2}{1.5 \, \text{m}}
\]
4. **Calculate**:
\[
\rho = 2 \times \frac{1 \times 10^{-6}}{1.5}
\]
\[
\rho = \frac{2 \times 10^{-6}}{1.5}
\]
\[
\rho = 1.33 \times 10^{-6} \, \Omega \cdot m
\]
5. **Check Units**: The units of resistivity are ohm-meters, which is correct.
### Common Tips
- **Units**: Always double-check the units of your given quantities. Convert them if necessary before substituting into the formula.
- **Consistency**: Ensure that the dimensions (length, area) are in the same unit system before substituting into the formula.
- **Accuracy**: Be careful with arithmetic operations and rounding to avoid calculation errors.
By following these steps, you should be able to solve most problems related to resistivity in Class 10.
### Key Concepts
1. **Resistivity (\(\rho\))**: Resistivity is a property of a material that indicates how much it resists the flow of electric current. It depends on the material and is defined by the equation:
\[
\rho = R \times \frac{A}{L}
\]
where \( R \) is the resistance, \( A \) is the cross-sectional area of the conductor, and \( L \) is the length of the conductor.
2. **Resistance (R)**: The resistance of a conductor is given by:
\[
R = \rho \times \frac{L}{A}
\]
Rearranging this equation gives us the resistivity:
\[
\rho = R \times \frac{A}{L}
\]
### Steps to Solve Numericals
1. **Understand the Problem**: Read the problem carefully to identify what is given and what needs to be found. Typically, you might be given values for resistance, length, and cross-sectional area and asked to find the resistivity, or vice versa.
2. **Write Down the Formula**: Based on what you need to find, write down the relevant formula. For resistivity, use:
\[
\rho = R \times \frac{A}{L}
\]
3. **Substitute the Values**: Plug in the given values into the formula. Make sure that the units are consistent. For example, if the length is given in meters, the area should be in square meters.
4. **Calculate**: Perform the arithmetic operations to solve for resistivity. Ensure that you carry out the calculations carefully to avoid mistakes.
5. **Check Units**: Ensure that the result has the correct units for resistivity, which is ohm-meters (\(\Omega \cdot m\)).
### Example Problem
**Problem**: A copper wire has a resistance of 2 ohms, a length of 1.5 meters, and a cross-sectional area of \(1 \times 10^{-6}\) square meters. Find the resistivity of copper.
**Solution**:
1. **Identify the Given Values**:
- Resistance \( R = 2 \, \Omega \)
- Length \( L = 1.5 \, \text{m} \)
- Cross-sectional Area \( A = 1 \times 10^{-6} \, \text{m}^2 \)
2. **Use the Resistivity Formula**:
\[
\rho = R \times \frac{A}{L}
\]
3. **Substitute the Values**:
\[
\rho = 2 \, \Omega \times \frac{1 \times 10^{-6} \, \text{m}^2}{1.5 \, \text{m}}
\]
4. **Calculate**:
\[
\rho = 2 \times \frac{1 \times 10^{-6}}{1.5}
\]
\[
\rho = \frac{2 \times 10^{-6}}{1.5}
\]
\[
\rho = 1.33 \times 10^{-6} \, \Omega \cdot m
\]
5. **Check Units**: The units of resistivity are ohm-meters, which is correct.
### Common Tips
- **Units**: Always double-check the units of your given quantities. Convert them if necessary before substituting into the formula.
- **Consistency**: Ensure that the dimensions (length, area) are in the same unit system before substituting into the formula.
- **Accuracy**: Be careful with arithmetic operations and rounding to avoid calculation errors.
By following these steps, you should be able to solve most problems related to resistivity in Class 10.