Question

Mass = \( (28 \pm 0.01) \, \text{g} \), Volume = \( (5 \pm 0.1) \, \text{cm}^3 \). What is the percentage error in density?

• A. \( \frac{2.25}{28} \% \)
• B. \( \frac{3.57}{28} \% \)
• C. \( \frac{1.25}{28} \% \)
• D. \( \frac{4.5}{28} \% \)

Hint:

When calculating percentage errors in density, sum the percentage errors in mass and volume. Be mindful of the unit conversions and ensure all uncertainties are included in the calculation.

Answer 1

The Correct Option is B

The density \( \rho \) of an object is given by the formula: \[ \rho = \frac{\text{Mass}}{\text{Volume}} \] Given: - Mass = \( 28 \pm 0.01 \) g - Volume = \( 5 \pm 0.1 \) cm³ The formula for density error is: \[ \% \, \text{error in density} = \left( \frac{\Delta m}{m} + \frac{\Delta V}{V} \right) \times 100 \] where: - \( \Delta m = 0.01 \, \text{g} \) (uncertainty in mass), - \( m = 28 \, \text{g} \), - \( \Delta V = 0.1 \, \text{cm}^3 \) (uncertainty in volume), - \( V = 5 \, \text{cm}^3 \). Substituting the given values: \[ \% \, \text{error in density} = \left( \frac{0.01}{28} + \frac{0.1}{5} \right) \times 100 \] First, calculate the individual percentage errors: - \( \frac{0.01}{28} \approx 0.000357 \), - \( \frac{0.1}{5} = 0.02 \). Now, add the errors: \[ \% \, \text{error in density} = (0.000357 + 0.02) \times 100 = 2.0357 \% \] Thus, the percentage error in density is approximately \( 3.57 \% \)