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 density of the wire is determined by the formula: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{\text{Mass}}{\pi r^2 L}, \] where \( r \) represents the radius and \( L \) represents the length of the wire. The maximum percentage error in the density is calculated as: \[ \left( \frac{\Delta \rho}{\rho} \right) = \left( \frac{\Delta m}{m} \right) + 2 \left( \frac{\Delta r}{r} \right) + \left( \frac{\Delta L}{L} \right), \] with \( \Delta m, \Delta r, \) and \( \Delta L \) denoting the absolute errors in mass, radius, and length, respectively. Step 1: Compute the individual 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: Sum the calculated errors. The cumulative percentage error in density is: \[ \frac{\Delta \rho}{\rho} = 0.5\% + 2(2\%) + 0.5\% = 5\%. \] Consequently, the maximum percentage error in the density measurement is \( \boxed{5} \).