Question

Each side of a metallic cube of mass 5.580 kg is measured to be 9.0 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as X $\times$ 10³ kg m⁻³, where the value of X is: ____.

• A. 7.654
• B. 7.6
• C. 7.65
• D. 7.7

Hint:

Don't get distracted by the high precision of the mass. The precision of your final answer is always limited by your "weakest link" — in this case, the side length measured to only two figures.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When performing multiplication or division, the final result should have the same number of significant figures as the measurement with the least number of significant figures.

Step 2: Key Formula or Approach:
\[ \text{Density } (\rho) = \frac{\text{Mass}}{\text{Volume}} = \frac{m}{s^3} \]

Step 3: Detailed Explanation:
1. Identify Significant Figures: - Mass ($m$) = 5.580 kg (4 significant figures) - Side ($s$) = 9.0 cm (2 significant figures) 2. Calculate Density: - Side $s = 0.090$ m - $\text{Volume} = (0.090)^3 = 0.000729 \text{ m}^3$ - $\rho = \frac{5.580}{0.000729} \approx 7654.3 \text{ kg/m}^3$ - $\rho \approx 7.6543 \times 10^3 \text{ kg/m}^3$ 3. Apply Rounding Rules: - Since the side (9.0) has only 2 significant figures, the final result must be rounded to 2 significant figures. - $7.6543 \dots$ rounded to two sig-figs is 7.7 (or 7.6 based on specific rounding rules for 5, but typically 7.7 as 5 is followed by non-zero digits). However, checking the provided options, we look for the 2 sig-fig representative.

Step 4: Final Answer:
Based on the 2 significant figure rule from the measurement "9.0 cm", the value of X is 7.7. (Note: Depending on exact arithmetic, 7.65... rounds up).