A cuboid with dimensions 4 cm, 3 cm, and 12 cm is three-quarters full of milk. If the weight of the milk is 2.5 N, calculate its density in kg/m³.

The dimensions of the cuboid are 4 cm, 3 cm, and 12 cm.

$$V_{cuboid} = \text{length} \times \text{width} \times \text{height}$$ $$V_{cuboid} = 4\, \text{cm} \times 3\, \text{cm} \times 12\, \text{cm}$$ $$V_{cuboid} = 144\, \text{cm}^3$$

The cuboid is three-quarter full of milk.

$$V_{milk} = \frac{3}{4} \times V_{cuboid}$$ $$V_{milk} = \frac{3}{4} \times 144\, \text{cm}^3$$ $$V_{milk} = 3 \times 36\, \text{cm}^3$$ $$V_{milk} = 108\, \text{cm}^3$$

We know that $1\, \text{m} = 100\, \text{cm}$, so $1\, \text{m}^3 = (100\, \text{cm})^3 = 1,000,000\, \text{cm}^3$.

$$V_{milk} = 108\, \text{cm}^3 \times \frac{1\, \text{m}^3}{1,000,000\, \text{cm}^3}$$ $$V_{milk} = 0.000108\, \text{m}^3$$

The weight of the milk ($W$) is 2.5 N. We use the formula $W = m \times g$, where $m$ is mass and $g$ is the acceleration due to gravity. We will use $g = 9.81\, \text{m/s}^2$.

$$m_{milk} = \frac{W}{g}$$ $$m_{milk} = \frac{2.5\, \text{N}}{9.81\, \text{m/s}^2}$$ $$m_{milk} \approx 0.25484\, \text{kg}$$

Density ($\rho$) is given by the formula $\rho = \frac{m}{V}$.

$$\rho_{milk} = \frac{m_{milk}}{V_{milk}}$$ $$\rho_{milk} = \frac{0.25484\, \text{kg}}{0.000108\, \text{m}^3}$$ $$\rho_{milk} \approx 2359.6\, \text{kg/m}^3$$

Rounding to three significant figures:

$$\rho_{milk} \approx 2360\, \text{kg/m}^3$$

The density of the milk is $\boxed{\text{2360 kg/m}^3}$.

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