Question

A metal wire has mass (0.4±0.002)g, radius (0.3±0.001)mm and length (5±0.02)cm. The maximum possible percentage error in the measurement of density will nearly be:

• A. 1.4%
• B. 1.2%
• C. 1.3%
• D. 1.6%

Answer 1

The Correct Option is D

To find the maximum possible percentage error in the measurement of the density of a metal wire, we can use the formula for density and the rules for error propagation. The density \(\rho\) of a metal wire can be calculated using the formula:

\(\rho = \frac{m}{V}\)

Where \(m\) is the mass and \(V\) is the volume. For a cylindrical wire, the volume \(V\) is given by:

\(V = \pi r^2 l\)

where:

• \(r\) is the radius,
• \(l\) is the length.

The formula for density is therefore:

\(\rho = \frac{m}{\pi r^2 l}\)

The percentage error in density, \(\Delta \rho / \rho \times 100\%\), using the rules of error propagation, is given by:

\(\frac{\Delta \rho}{\rho} \times 100\% = \left( \frac{\Delta m}{m} + 2 \frac{\Delta r}{r} + \frac{\Delta l}{l} \right) \times 100\%\)

Given measurements and their errors:

• Mass, \(m = 0.4\, \text{g}\) with \(\Delta m = 0.002\, \text{g}\)
• Radius, \(r = 0.3\, \text{mm} = 0.03\, \text{cm}\) with \(\Delta r = 0.001\, \text{mm} = 0.0001\, \text{cm}\)
• Length, \(l = 5\, \text{cm}\) with \(\Delta l = 0.02\, \text{cm}\)

Now, calculate each fractional error:

1. \(\frac{\Delta m}{m} = \frac{0.002}{0.4} = 0.005\quad \text{or}\quad 0.5\%\)
2. \(\frac{\Delta r}{r} = \frac{0.0001}{0.03} \approx 0.00333\quad \text{or}\quad 0.333\%\)
3. \(\frac{\Delta l}{l} = \frac{0.02}{5} = 0.004\quad \text{or}\quad 0.4\%\)

Incorporating these into the error propagation formula gives the maximum percentage error in density:

\(\frac{\Delta \rho}{\rho} \approx \left(0.005 + 2 \times 0.00333 + 0.004\right) \times 100\% \approx 0.016\quad \text{or}\quad 1.6\%\)

Therefore, the maximum possible percentage error in the measurement of the density is 1.6%.