The radius (\(r\)), length (\(l\)), and resistance (\(R\)) of a metal wire were measured in the laboratory as: \(r = (0.35 \pm 0.05) \, \text{cm}, \quad R = (100 \pm 10) \, \text{ohm}, \quad l = (15 \pm 0.2) \, \text{cm}.\)
The percentage error in the resistivity of the material of the wire is:
The percentage error in the resistivity of the material of the wire is:
- \( 25.6\% \)
- \( 39.9\% \)
- \( 37.3\% \)
- \( 35.6\% \)
The Correct Option is B
Approach Solution - 1
The formula for resistivity is given by:
\(\rho = R \frac{\pi r^2}{l}\)
The relative error in resistivity is given by:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}\)
Substituting the given values:
\(\frac{\Delta \rho}{\rho} = \frac{10}{100} + 2 \times \frac{0.05}{0.35} + \frac{0.2}{15}\)
Calculating each term:
\(\frac{\Delta \rho}{\rho} = 0.1 + 2 \times 0.1429 + 0.0133\)
\(\frac{\Delta \rho}{\rho} \approx 0.1 + 0.2858 + 0.0133 = 0.3991 \approx 39.9\%\)
Approach Solution -2
The problem requires us to calculate the percentage error in the resistivity of a metal wire, given the measured values and their associated errors for the wire's radius, length, and resistance.
Concept Used:
1. Formula for Resistivity: The resistance \(R\) of a wire is related to its resistivity \(\rho\), length \(l\), and cross-sectional area \(A\) by the formula:
\[ R = \rho \frac{l}{A} \]
For a wire with a circular cross-section of radius \(r\), the area is \(A = \pi r^2\). Substituting this into the resistance formula and solving for resistivity \(\rho\), we get:
\[ \rho = \frac{R (\pi r^2)}{l} \]
2. Propagation of Errors: For a quantity \(Z\) calculated from measured quantities \(X\), \(Y\), and \(W\) using a formula of the form \(Z = k \cdot X^a \cdot Y^b \cdot W^c\), where \(k\) is a constant, the fractional error in \(Z\) is given by the sum of the fractional errors in the measured quantities, each multiplied by the magnitude of its power:
\[ \frac{\Delta Z}{Z} = a \frac{\Delta X}{X} + b \frac{\Delta Y}{Y} + c \frac{\Delta W}{W} \]
The percentage error is then \( \left( \frac{\Delta Z}{Z} \right) \times 100\% \).
Step-by-Step Solution:
Step 1: List the given measurements and their absolute errors.
Radius, \(r = 0.35 \, \text{cm}\) with absolute error \(\Delta r = 0.05 \, \text{cm}\).
Resistance, \(R = 100 \, \text{ohm}\) with absolute error \(\Delta R = 10 \, \text{ohm}\).
Length, \(l = 15 \, \text{cm}\) with absolute error \(\Delta l = 0.2 \, \text{cm}\).
Step 2: Write the formula for resistivity and derive the expression for its fractional error.
The formula for resistivity is \(\rho = \frac{R \pi r^2}{l}\). Applying the rule for propagation of errors (note that the constant \(\pi\) has no error associated with it):
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \left( \frac{\Delta r}{r} \right) + \frac{\Delta l}{l} \]
This equation gives the total fractional error in resistivity.
Step 3: Calculate the individual fractional errors for \(R\), \(r\), and \(l\).
Fractional error in resistance:
\[ \frac{\Delta R}{R} = \frac{10}{100} = 0.1 \]
Fractional error in radius:
\[ \frac{\Delta r}{r} = \frac{0.05}{0.35} = \frac{5}{35} = \frac{1}{7} \]
Fractional error in length:
\[ \frac{\Delta l}{l} = \frac{0.2}{15} = \frac{2}{150} = \frac{1}{75} \]
Step 4: Substitute these fractional errors into the formula for the total fractional error in resistivity.
\[ \frac{\Delta \rho}{\rho} = 0.1 + 2 \left( \frac{1}{7} \right) + \frac{1}{75} \]
Step 5: Compute the numerical value of the total fractional error.
\[ \frac{\Delta \rho}{\rho} \approx 0.1 + 2(0.1428) + 0.0133 \] \[ \frac{\Delta \rho}{\rho} \approx 0.1 + 0.2857 + 0.0133 \] \[ \frac{\Delta \rho}{\rho} \approx 0.399 \]
Step 6: Convert the fractional error to a percentage error by multiplying by 100.
\[ \text{Percentage Error in } \rho = \left( \frac{\Delta \rho}{\rho} \right) \times 100\% \] \[ \text{Percentage Error} \approx 0.399 \times 100\% = 39.9\% \]
Rounding to the nearest whole number, the percentage error is approximately 39.9%.
Therefore, the percentage error in the resistivity of the material is 39.9%.
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