Question

A cylindrical wire has a mass(0.3 ± 0.003)g, radius (0.5 ± 0.005)mm and length (6 ± 0.06)cm. The maximum percentage error in the measurement of its density is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is D

We are given the measurements of a cylindrical wire with their uncertainties:

• Mass ($m$) = $0.3 \pm 0.003 \, \text{g}$
• Radius ($r$) = $0.5 \pm 0.005 \, \text{mm}$
• Length ($L$) = $6 \pm 0.06 \, \text{cm}$

We need to find the maximum percentage error in the measurement of its density ($\rho$).

Key Concepts:

1. Density Formula: Density is mass per unit volume ($\rho = \frac{m}{V}$).
2. Volume of a Cylinder: The volume ($V$) of a cylinder is $V = \pi r^2 L$.
3. Density in terms of measured quantities: Substituting the volume formula, $\rho = \frac{m}{\pi r^2 L}$.
4. Propagation of Errors (Maximum Percentage Error): For a quantity $Q = \frac{x^a y^b}{z^c}$, the maximum percentage error in $Q$ is given by:

$\left( \frac{\Delta Q}{Q} \times 100\% \right)_{\text{max}} = \left( |a| \frac{\Delta x}{x} + |b| \frac{\Delta y}{y} + |c| \frac{\Delta z}{z} \right) \times 100\%$

1. Applying this to the density formula $\rho = \frac{m^1}{\pi r^2 L^1}$ (note $\pi$ is a constant with no error), the maximum percentage error in density is:

$\left( \frac{\Delta \rho}{\rho} \times 100\% \right)_{\text{max}} = \left( (1) \frac{\Delta m}{m} + (2) \frac{\Delta r}{r} + (1) \frac{\Delta L}{L} \right) \times 100\%$

Calculations:

First, calculate the individual percentage errors for mass, radius, and length:

1. Percentage error in mass ($\frac{\Delta m}{m} \times 100\%$):

$\frac{0.003 \, \text{g}}{0.3 \, \text{g}} \times 100\% = \frac{3}{300} \times 100\% = \frac{1}{100} \times 100\% = 1\%$

1. Percentage error in radius ($\frac{\Delta r}{r} \times 100\%$):

$\frac{0.005 \, \text{mm}}{0.5 \, \text{mm}} \times 100\% = \frac{5}{500} \times 100\% = \frac{1}{100} \times 100\% = 1\%$

1. Percentage error in length ($\frac{\Delta L}{L} \times 100\%$):

$\frac{0.06 \, \text{cm}}{6 \, \text{cm}} \times 100\% = \frac{6}{600} \times 100\% = \frac{1}{100} \times 100\% = 1\%$

Now, substitute these percentage errors into the formula for the percentage error in density:

$\left( \frac{\Delta \rho}{\rho} \times 100\% \right)_{\text{max}} = \left( \frac{\Delta m}{m} \times 100\% \right) + 2 \left( \frac{\Delta r}{r} \times 100\% \right) + \left( \frac{\Delta L}{L} \times 100\% \right)$

$\left( \frac{\Delta \rho}{\rho} \times 100\% \right)_{\text{max}} = (1\%) + 2 \times (1\%) + (1\%)$

$\left( \frac{\Delta \rho}{\rho} \times 100\% \right)_{\text{max}} = 1\% + 2\% + 1\% = 4\%$

The maximum percentage error in the measurement of the density is 4%.

The final answer is ${4}$

Answer 2

The density \( \rho \) of a cylindrical wire is given by the formula: \[ \rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m}{\pi r^2 L} \] Where: - \( m \) is the mass, - \( r \) is the radius, - \( L \) is the length. The percentage error in the density is given by the sum of the percentage errors in mass, radius, and length: \[ \% \, \text{error in} \, \rho = \% \, \text{error in} \, m + 2 \times \% \, \text{error in} \, r + \% \, \text{error in} \, L \] Where: - The factor of 2 comes from the \( r^2 \) term in the formula for volume. Now, let’s calculate the percentage errors: - The mass \( m = 0.3 \, \text{g} \), and the error in mass is \( 0.003 \, \text{g} \). So the percentage error in mass is: \[ \% \, \text{error in} \, m = \frac{0.003}{0.3} \times 100 = 1\% \] - The radius \( r = 0.5 \, \text{mm} \), and the error in radius is \( 0.005 \, \text{mm} \). So the percentage error in radius is: \[ \% \, \text{error in} \, r = \frac{0.005}{0.5} \times 100 = 1\% \] - The length \( L = 6 \, \text{cm} \), and the error in length is \( 0.06 \, \text{cm} \). So the percentage error in length is: \[ \% \, \text{error in} \, L = \frac{0.06}{6} \times 100 = 1\% \] Now, summing up the errors: \[ \% \, \text{error in} \, \rho = 1 + 2 \times 1 + 1 = 4\% \] Thus, the maximum percentage error in the measurement of the density is \( 4\% \).