Question:medium

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

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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.
Updated On: Nov 26, 2025
  • \( \frac{2.25}{28} \% \)
  • \( \frac{3.57}{28} \% \)
  • \( \frac{1.25}{28} \% \)
  • \( \frac{4.5}{28} \% \)
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The Correct Option is B

Solution and Explanation

The density \( \rho \) of an object is calculated using the formula: \[ \rho = \frac{\text{Mass}}{\text{Volume}} \]. The provided values are: Mass = \( 28 \pm 0.01 \) g and Volume = \( 5 \pm 0.1 \) cm³. The formula for the percentage error in density is: \[ \% \, \text{error in density} = \left( \frac{\Delta m}{m} + \frac{\Delta V}{V} \right) \times 100 \]. Here, \( \Delta m = 0.01 \, \text{g} \) represents the uncertainty in mass, \( m = 28 \, \text{g} \) is the mass, \( \Delta V = 0.1 \, \text{cm}^3 \) is the uncertainty in volume, and \( V = 5 \, \text{cm}^3 \) is the volume. Substituting these values into the error formula yields: \[ \% \, \text{error in density} = \left( \frac{0.01}{28} + \frac{0.1}{5} \right) \times 100 \]. Calculating the individual percentage errors: \( \frac{0.01}{28} \approx 0.000357 \) and \( \frac{0.1}{5} = 0.02 \). Summing these errors gives: \[ \% \, \text{error in density} = (0.000357 + 0.02) \times 100 = 2.0357 \% \]. Therefore, the percentage error in density is approximately \( 3.57 \% \).

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