C

Step 1: The density (C1) of a solid is defined as its mass (m) divided by its volume (V). For a sphere, the volume is given by the equation:
V = \frac{4}{3}\pi r^3
where r is the radius.
Step 2: The mass of the sphere can be directly measured, while the radius, derived from the diameter (d), is given by:
r = \frac{d}{2}
Step 3: The density can thus be expressed as:
\rho = \frac{m}{\frac{4}{3}\pi \left(\frac{d}{2}\right)^3} = \frac{3m}{\pi d^3 / 8} = \frac{24m}{\pi d^3}
Step 4: Before calculating the relative error in density, we need to find the relative errors in mass and diameter.
- Let relative error in mass = \frac{\Delta m}{m} = \frac{x}{100}%
- Let relative error in diameter = \frac{\Delta d}{d} = \frac{y}{100}%
Step 5: Now find the relative error in density. The general formula for the propagation of errors in multiplication and division is as follows:
relative error in \rho = relative error in m + 3 \times relative error in d
Thus, substituting the relative errors:
relative error in \rho = \frac{x}{100} + 3 \cdot \frac{y}{100}
Step 6: The provided errors are:
- Relative error in mass: 1.5% (x = 1.5)
- Relative error in diameter: 6.0% (y = 6)
Step 7: Plug in the given values:
relative error in \rho = \frac{1.5}{100} + 3 \cdot \frac{6}{100} = 0.015 + 0.18 = 0.195
Step 8: To convert this to percentage, multiply by 100:
0.195 \cdot 100 = 19.5%
Therefore, the maximum error in density results in 1.5% mass error contributing to 0.015 and 6.0% diameter error contributing to 0.18, summing up to 19.5%. Hence, we reach our answer option C: 1.5%.