The density of a solid metal sphere is diameter. The maximum error in the density of the sphere is $$\left(\frac{x}{100}\right)\%$$. If the relative errors in measuring the mass and the diameter are $$6.0\%$$ and $$1.5\%$$ respectively, the value of $$x$$ is ___

We have to determine the maximum percentage error in the density of a solid metal sphere when its mass and diameter are measured with known percentage (relative) errors.

First, we recall the basic definition of density.

Density $$\rho$$ is defined by the formula

$$\rho \;=\; \frac{\text{mass}}{\text{volume}} \;=\; \frac{m}{V}.$$

For a sphere whose diameter is $$d$$, the radius is $$r = \dfrac{d}{2}$$. The volume of a sphere is

$$V \;=\; \frac{4}{3}\pi r^{3}.$$

Substituting $$r = \dfrac{d}{2}$$, we find

$$V \;=\; \frac{4}{3}\pi\left(\frac{d}{2}\right)^{3} \;=\; \frac{4}{3}\pi\frac{d^{3}}{8} \;=\; \frac{\pi d^{3}}{6}.$$

Now we write the density explicitly in terms of the mass $$m$$ and the diameter $$d$$.

$$\rho \;=\; \frac{m}{\dfrac{\pi d^{3}}{6}} \;=\; \frac{6m}{\pi d^{3}}.$$

Hence $$\rho$$ is proportional to $$m$$ and inversely proportional to $$d^{3}$$. In algebraic form we may write

$$\rho \;\propto\; m^{\;1}\,d^{-3}.$$

To find the maximum (worst-case) percentage error in $$\rho$$, we use the well-known rule for propagation of relative errors for products and quotients:

When a quantity $$Q$$ depends on measured quantities as $$Q = A^{\,\alpha}B^{\,\beta}$$, the maximum relative error in $$Q$$ is

$$\frac{\Delta Q}{Q} \;=\; |\alpha|\frac{\Delta A}{A} \;+\; |\beta|\frac{\Delta B}{B}.$$

Applying this rule to $$\rho \propto m^{1}d^{-3}$$, we obtain

$$\frac{\Delta\rho}{\rho} \;=\; 1\bigl(\frac{\Delta m}{m}\bigr) + 3\bigl(\frac{\Delta d}{d}\bigr).$$

The problem states that the percentage (relative) errors in measuring the mass and the diameter are respectively

$$\frac{\Delta m}{m} = 6.0\%, \qquad \frac{\Delta d}{d} = 1.5\%.$$

Substituting these values, we get

$$\frac{\Delta\rho}{\rho} \;=\; 1(6.0\%) + 3(1.5\%) \;=\; 6.0\% + 4.5\% \;=\; 10.5\%.$$

According to the wording of the question, this maximum error is written in the form

$$\left(\frac{x}{100}\right)\%.$$

Equating the two expressions for the percentage error, we have

$$\left(\frac{x}{100}\right)\% = 10.5\%.$$

Removing the percent sign (which merely indicates “per hundred”) and equating the numerical parts, we find

$$\frac{x}{100} = 10.5 \;\;\Longrightarrow\;\; x = 10.5 \times 100 = 1050.$$

So, the answer is $$1050$$.