A solid sphere with a radius of 5 cm floats in water. If a maximum...

Let the volume of the sphere be $V$ and the density of its material be $\rho_s$.

Volume of the sphere is given by:

$V = \dfrac{4}{3} \pi R^3$

Substituting $R = 5 \text{ cm}$:

$V = \dfrac{4}{3} \pi (5)^3 = \dfrac{500 \pi}{3} \text{ cm}^3$

When the maximum load is placed on the sphere without getting wet, the sphere is completely submerged in water. By Archimedes' principle, the buoyant force equals the total weight of the sphere and the load.

$V \rho_w g = (M + m) g$

$V \rho_w = V \rho_s + m$

Given the mass of the load $m = 0.1 \text{ kg} = 100 \text{ g}$ and the density of water $\rho_w = 1 \text{ g/cm}^3$:

$\dfrac{500 \pi}{3} (1) = \dfrac{500 \pi}{3} \rho_s + 100$

Dividing the equation by $\dfrac{500 \pi}{3}$:

$1 = \rho_s + \dfrac{300}{500 \pi}$

$\rho_s = 1 - \dfrac{3}{5 \pi}$

Approximating $\pi \approx 3.14$:

$\rho_s = 1 - \dfrac{3}{15.7} \approx 1 - 0.191 = 0.809 \text{ g/cm}^3$

The specific gravity of the sphere's material is numerically equal to its density in $\text{g/cm}^3$, which is approximately $0.8$.