Questions: A student measures the sides of a cube as 5 cm and the mass as 124 g. The true density of the material is 1.02 g / cm^3. Calculate the percent error in the student's measurement.

Solution Steps

The volume \( V \) of a cube is calculated using the formula: \[ V = \text{side}^3 \] Given that the side of the cube is 5 cm, the volume is: \[ V = 5^3 = 125 \, \text{cm}^3 \]

The density \( \rho \) is calculated using the formula: \[ \rho = \frac{\text{mass}}{\text{volume}} \] The student measured the mass as 124 g, so the measured density is: \[ \rho = \frac{124 \, \text{g}}{125 \, \text{cm}^3} = 0.9920 \, \text{g/cm}^3 \]

The percent error is calculated using the formula: \[ \text{Percent Error} = \left| \frac{\text{True Value} - \text{Measured Value}}{\text{True Value}} \right| \times 100\% \] The true density is \(1.02 \, \text{g/cm}^3\), so the percent error is: \[ \text{Percent Error} = \left| \frac{1.02 - 0.9920}{1.02} \right| \times 100\% = \left| \frac{0.0280}{1.02} \right| \times 100\% \approx 2.7451\% \]

Final Answer