The density of a material, in the shape of a cube, is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are 1.5% and 1%, respectively, the maximum error in determining the density is:

We have to find the percentage (relative) error in the density $$\rho$$ of a cubical block. By definition,

$$\rho=\frac{\text{mass}}{\text{volume}}$$

For a cube of side $$l$$, the volume is $$l^{3}$$. Hence,

$$\rho=\frac{m}{l^{3}}$$

Now, we apply the general rule for propagation of relative errors in a product or quotient. The rule states:

If $$Q = a^{p}\,b^{q}\,c^{r}\ldots$$, then the relative error in $$Q$$ is given by

$$\frac{\Delta Q}{Q}=|p|\,\frac{\Delta a}{a}+|q|\,\frac{\Delta b}{b}+|r|\,\frac{\Delta c}{c}+\ldots$$

Comparing our expression $$\rho = m^{1}\,l^{-3}$$ with the rule, we see that the exponent of $$m$$ is $$+1$$ and the exponent of $$l$$ is $$-3$$. Taking absolute values of these exponents, we write

$$\frac{\Delta \rho}{\rho} = 1\;\frac{\Delta m}{m} + 3\;\frac{\Delta l}{l}$$

The problem tells us that the relative error in measuring the mass is $$1.5\%$$. Therefore,

$$\frac{\Delta m}{m}=1.5\%$$

Similarly, the relative error in each length measurement is $$1\%$$, so

$$\frac{\Delta l}{l}=1\%$$

Substituting these values into the error‐propagation formula gives

$$\frac{\Delta \rho}{\rho}=1\,(1.5\%) + 3\,(1\%)$$

$$\frac{\Delta \rho}{\rho}=1.5\% + 3\%$$

$$\frac{\Delta \rho}{\rho}=4.5\%$$

This value represents the maximum possible percentage error in the calculated density.

Hence, the correct answer is Option D.