Answer:
a) [tex]m_{sph} = 3.768\,kg[/tex], b) [tex]\rho_{sph} = 4897.795\,\frac{kg}{m^{3}}[/tex]
Explanation:
According to the Archimedes' Principle, the buoyancy force is equal the weight of the displaced fluid. Then, this equation of equilibrium is constructed by the Newton's Laws:
[tex]\Sigma F = \rho_{w}\cdot V_{disp}\cdot g -\rho_{sph}\cdot V_{sph}\cdot g = 0[/tex]
After some algebraic handling:
[tex]\rho_{w}\cdot V_{disp} = \rho_{sph}\cdot V_{sph}[/tex]
a) The mass of the sphere is:
[tex]m_{sph} = \rho_{w}\cdot V_{disp}[/tex]
[tex]m_{sph} = (927\,\frac{kg}{m^{3}} )\cdot \left[\frac{4}{3}\pi\cdot (0.099\,m)^{3} \right][/tex]
[tex]m_{sph} = 3.768\,kg[/tex]
b) The density of the sphere is:
[tex]\rho_{sph} = \frac{m_{sph}}{V_{sph}}[/tex]
[tex]\rho_{sph} = \frac{3.768\,kg}{\frac{4}{3}\pi\cdot [(0.099\,m)^{3}-(0.087\,m)^{3}]}[/tex]
[tex]\rho_{sph} = 4897.795\,\frac{kg}{m^{3}}[/tex]
