Question
Each side of a metallic cube of mass 5.580 kg is measured to the 9.0 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as $X \times 10^{3}\text{ }kg{\text{ }m}^{- 3}$ where the value of $X$ is:
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Solution
Sol. [Key Concept: Significant figures & density = m/a³] Density $= \frac{\text{~}\text{Mass}\text{~}}{\text{~}\text{Volume}\text{~}}$
$${\begin{matrix} & \ = \frac{5.580\text{ }kg}{(9.0\text{ }cm)^{3}} \\ & \ = \frac{5.580}{729} \times 10^{6}\frac{\text{ }kg}{{\text{ }m}^{3}} \\ & \ = 0.007654321 \times 10^{6}\frac{\text{ }kg}{{\text{ }m}^{3}} \\ & \ = 7.654 \times 10^{3}\frac{\text{ }kg}{{\text{ }m}^{3}} \\ & \ = 7.7 \times 10^{3}\frac{\text{ }kg}{{\text{ }m}^{3}} \end{matrix} } {Hence,\ on\ comparing } {X = 7.7}$$
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