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Homework Statement:
 A solid sphere of volume 20 cm and uniform density ##\rho_B## = 400 ##\text{kg m}^{3}## floats in a vessel containing pure water. Calculate the depth to which the sphere is submerged.
Relevant Equations:

1. ##\text{Law of floatation}## : When a body floats in a liquid, the weight of the body is equal to that of the weight of the liquid displaced : ##w_B = \Delta w_L##.
2. The volume of a spherical body of radius ##r_B## is given by : ##V_B = \frac{4}{3}\pi r_B^3##.
3. The weight of a body is given by : ##w_B = m_B g = \rho_B V_B g## where ##\rho_B## is the density of the body and ##V_B,m_B## are the volume and mass of the body respectively.
The body has a radius of ##r_B = 0.2## m.
The volume of the body is given by ##V_B = \frac{4}{3}\times \pi \times 0.2^3 = 0.034\;\text{m}^3##.
The mass of the body : ##w_B = \rho_B V_B = 400\times 0.034 = 13.37\,\text{kg}##.
Hence the mass of the liquid displaced : ##\Delta m_L = 13.37\, \text{kg}## . (Law of floatation can also be "read off" as ##m_B = \Delta m_L##)
The volume of liquid displaced : ##\Delta V_L = \frac{\Delta m_L}{\rho_L} = \frac{13.37}{1000} = 1.34\times 10^{3}\,\text{m}^3##.
This is the volume of the sphere below the blue line inside the water, as shown to the right.
Question is, how to connect the volume of liquid displaced ##\mathbf{\Delta V_L}## to the volume of the body ##\mathbf{V_B}## ?
I could proceed no further. Any help would be appreciated.