Question

The maximum percentage error in the measurement of the density of a wire is: \[ \text{Given, mass of wire} = (0.60 \pm 0.003) \, \text{g}, \quad \text{radius of wire} = (0.50 \pm 0.01) \, \text{cm}, \quad \text{length of wire} = (10.00 \pm 0.05) \, \text{cm}. \]

• A. 7
• B. 5
• C. 4
• D. 8

Hint:

When calculating percentage errors for derived quantities: - For a quantity of the form \( A = \frac{m}{r^2 L} \), the percentage error is the sum of the percentage errors in each variable, with appropriate powers. - For example, the radius \( r^2 \) contributes twice the percentage error in radius.

Answer 1

The Correct Option is B

The formula for the density of the wire is: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{\text{Mass}}{\pi r^2 L}, \] where \( r \) is the radius and \( L \) is the length of the wire. The maximum percentage error in the density is given by: \[ \left( \frac{\Delta \rho}{\rho} \right) = \left( \frac{\Delta m}{m} \right) + 2 \left( \frac{\Delta r}{r} \right) + \left( \frac{\Delta L}{L} \right), \] where \( \Delta m, \Delta r, \) and \( \Delta L \) are the absolute errors in mass, radius, and length respectively. 

Step 1: Calculate the percentage errors. - Percentage error in mass: \( \frac{\Delta m}{m} = \frac{0.003}{0.60} \times 100 = 0.5\% \), - Percentage error in radius: \( \frac{\Delta r}{r} = \frac{0.01}{0.50} \times 100 = 2\% \), - Percentage error in length: \( \frac{\Delta L}{L} = \frac{0.05}{10.00} \times 100 = 0.5\% \). 

Step 2: Add the errors. The total percentage error in density: \[ \frac{\Delta \rho}{\rho} = 0.5\% + 2(2\%) + 0.5\% = 5\%. \] 

Thus, the maximum percentage error in the measurement of the density is \( \boxed{5} \).