To determine the density of a solid sphere, its mass is measured as...

The least count of the screw gauge is given by:
$\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of circular scale divisions}} = \dfrac{1\text{ mm}}{100} = 0.01\text{ mm}$

This least count represents the absolute error in the measurement of the diameter, so $\Delta D = 0.01\text{ mm}$.

The absolute error in mass is the least count of the weighing instrument, so $\Delta m = 0.01\text{ g}$.

The density $\rho$ of a sphere is given by:
$\rho = \dfrac{m}{V} = \dfrac{m}{\dfrac{4}{3}\pi \left(\dfrac{D}{2}\right)^3} = \dfrac{6m}{\pi D^3}$

The maximum relative error in density is:
$\dfrac{\Delta \rho}{\rho} = \dfrac{\Delta m}{m} + 3\dfrac{\Delta D}{D}$

Percentage error in mass $m$ is:
$\dfrac{\Delta m}{m} \times 100 = \dfrac{0.01}{10.00} \times 100 = 0.1\%$

Percentage error in diameter $D$ is:
$\dfrac{\Delta D}{D} \times 100 = \dfrac{0.01}{2.00} \times 100 = 0.5\%$

Therefore, the maximum percentage error in density is:
\left(\dfrac{\Delta \rho}{\rho} \times 100\right) = 0.1\% + 3(0.5\%) = 0.1\% + 1.5\% = 1.6\%$<br><br>Answer:$1.6\%