Problem 77 - Master Resource Book in JEE Main Physics

Question

The specific resistance \(\rho\) of a circular wire of radius \(r_{1}\) resistance \(R\) and length \(l\) is given by \(\rho=\frac{\pi r^{2} R}{l}\). Given, \(r=0:(24 \pm 0.02) \mathrm{cm}, R=(30 \pm 1) \Omega\) and \(l=(4.80 \pm 0.01) \mathrm{cm}\). The percentage error in \(\rho\) is nearly (a) \(7 \%\) (b) \(9 \%\) (c) 139 (d) \(20 \%\)

Step-by-Step Solution

Answer

The percentage error in \( \rho \) is nearly 20\% (option d).

Step 1: Write the formula for specific resistance

The specific resistance \( \rho \) of a wire is given by the formula: \[ \rho = \frac{\pi r^2 R}{l} \] where \( r \) is the radius, \( R \) is the resistance, and \( l \) is the length of the wire.

Step 2: Calculate the relative errors

Relative error for each quantity is given by: \[ \frac{\Delta x}{x} \] - For the radius \( r \): \( \frac{\Delta r}{r} = \frac{0.02}{0.24} \approx 0.0833 \) - For the resistance \( R \): \( \frac{\Delta R}{R} = \frac{1}{30} \approx 0.0333 \) - For the length \( l \): \( \frac{\Delta l}{l} = \frac{0.01}{4.80} \approx 0.0021 \) Note these values are in decimal form, convert to percentages by multiplying by 100.

Step 3: Find the formula for percentage error in \( \rho \)

Use the general rule of error propagation for the product or quotient of measurements:\[ \frac{\Delta \rho}{\rho} = 2 \frac{\Delta r}{r} + \frac{\Delta R}{R} + \frac{\Delta l}{l} \]. This involves adding up the relative errors, and the \( \Delta r/r \) term is multiplied by 2 because the radius \( r \) is squared in the formula.

Step 4: Calculate the percentage error in \( \rho \)

Substitute the relative errors: - \( 2 \frac{\Delta r}{r} = 2 \times 0.0833 = 0.1666 \) - \( \frac{\Delta R}{R} = 0.0333 \) - \( \frac{\Delta l}{l} = 0.0021 \) Sum these up to find the total percentage error: \[ \frac{\Delta \rho}{\rho} = 0.1666 + 0.0333 + 0.0021 \approx 0.202 \]Convert to a percentage: \( 0.202 \times 100 \approx 20.2\% \).

Step 5: Choose the nearest percentage error option

The closest option to the calculated percentage error of approximately \( 20.2\% \) is \( 20\% \). Therefore, the correct answer is option (d) \( 20\% \).

Key Concepts

Error Propagation

Error propagation is essential in physics and engineering to understand how uncertainties in measurements affect the outcome. When we calculate a derived quantity using measured values, we need to account for the errors in those initial measurements. This ensures that our results are as accurate as possible.

The principle of error propagation involves determining how the uncertainties in individual measurements combine to affect the uncertainty in the final calculated value. In our case of the specific resistance \[ \rho = \frac{\pi r^2 R}{l} \],we need to consider the errors in radius \( r \), resistance \( R \), and length \( l \).

Since the radius appears squared in the formula, the error in \( r \) must be multiplied by 2. This is a crucial point: whenever a variable is raised to a power in your calculation, multiply its relative error by that power.

• For a radius \( r \), to find its effect on the final result, the contribution to the total relative error is \( 2 \frac{\Delta r}{r} \).
• This method makes it easier to visualize how each measurement impacts the result.

Knowing this rule helps to mitigate uncertainty across all areas where measurements are combined mathematically. This allows us to report our final results, like specific resistance, with confidence in their accuracy.

Electrical Resistance

Electrical resistance is a fundamental concept in electricity that quantifies how much an object opposes the flow of electric current. When working with circuits, knowing the resistance of components like wires helps in designing and understanding their behavior.

It is measured in ohms (\( \Omega \)) and represented by the symbol \( R \). Resistance can be influenced by several factors such as the material of the wire, its length, and its cross-sectional area.

• The resistance \( R \) of a wire can be calculated using resistance formulae and is often part of more complex calculations, like in the specific resistance formula.
• Understanding how to apply these calculations in real-world scenarios helps in both academic and practical applications, such as predicting how circuits will perform.

The core idea is, higher resistance means less current can flow through, given the same voltage. This principle helps in controlling and managing electric circuits and devices.

Relative Error Calculation

Relative error calculation is a method used to express the uncertainty of a measurement in relation to its size. Unlike absolute error, which gives us an idea of the error in the same units as the measurement, relative error provides perspective on how significant the error is.

To calculate the relative error, you divide the absolute error (the uncertainty) by the measured value itself. Then, for practical understanding, you often convert this to a percentage.

• For a measurement \( x \), relative error is calculated as \( \frac{\Delta x}{x} \).
• This step is vital because it informs us how much error is present per unit of measurement, which is crucial in situations requiring high precision.

In the context of specific resistance calculation, we calculate relative errors for the radius, resistance, and length, then use these to find the relative error in \( \rho \).

• By summing up each of these percentage errors, we get an overall idea of the error in the final derived measurement.
• When presented as a percentage, it communicates clearly whether our measurements are sufficiently precise or not.

Recognizing that relative error provides useful insight, students can better prepare for lab situations and data analysis by evaluating the reliability and quality of their measurements.