#### Answer

The maximum mass that can hang from the sphere is 55.6 kg

#### Work Step by Step

We find the volume of the sphere.
$V = \frac{4}{3}\pi~r^3$
$V = \frac{4}{3}\pi~(0.25~m)^3$
$V = 0.06545~m^3$
We find the mass $M_s$ of the sphere.
$M_s = \rho~V$
$M_s = (150~kg/m^3)(0.06545~m^3)$
$M_s = 9.818~kg$
The buoyant force is equal to the weight of the water that is displaced by the sphere's volume. Let $\rho_w$ be the density of water. We can find the buoyant force on the sphere when the whole sphere is submerged.
$F_B = \rho_w~V~g$
$F_B = (1000~kg/m^3)(0.06545~m^3)(9.80~m/s^2)$
$F_B = 641.4~N$
To find the maximum mass $M_o$ of an object that can hang on the sphere, we can assume that the whole sphere is submerged and the buoyant force is equal to the total weight of the sphere and the object.
$M_s~g+M_o~g = F_B$
$M_o = \frac{F_B-M_s~g}{g}$
$M_o = \frac{(641.4~N)-(9.818~kg)(9.80~m/s^2)}{9.80~m/s^2}$
$M_o = 55.6~kg$
The maximum mass that can hang from the sphere is 55.6 kg.