# Problem No. 6 (10 points)

The density of a body is defined as the ratio of its mass to the volume it occupies. A homogeneous cube with a volume of \( V = 27 \partial \mu^{3} \) is given. As a result of heating, each of its edges increased by 9 mm. By what percentage did the density of this cube change?
# Answer: decreased by $8 \%$

## Solution and evaluation criteria:

Volume of a cube: $v=a^{3}$, where $a$ is the length of the edge, therefore:

$a=3 dm=300 mm$.

Final edge length: $a_{K}=309 mm$.

Thus, the final volume: $V_{K}=a_{K}^{3}=3.09^{3}=29.503629 dm^{3} \approx 1.09 V$.

Therefore, the density:

$\rho_{K}=\frac{m}{V_{K}}=\frac{m}{1.09 V}=0.92 \frac{m}{V} \approx 0.92 \rho$. The density has decreased by approximately $8 \%$. (4 points)

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