Problem Analysis
We are lifting a sphere of mass $m$ and radius $r$ from the bottom of a reservoir. A crucial piece of information is that the density of the sphere is equal to the density of water ($\rho_s = \rho_w$).
Step 1: Work Done While Submerged
Since the density of the sphere equals the density of water ($\rho_s = \rho_w$), the sphere is neutrally buoyant. The Apparent Weight ($W_{app}$) inside water is:
$$W_{app} = mg – F_B = mg – (\rho_w V g)$$Since $m = \rho_s V$ and $\rho_s = \rho_w$, then $mg = \rho_w V g$.
$$W_{app} = 0$$Therefore, the force required to lift the sphere from the bottom to the surface (while fully submerged) is zero. Work done in this phase is 0.
Step 2: Work Done During Exit
Work is done only as the sphere crosses the surface. This process is energetically equivalent to lifting a mass $m$ from a state of weightlessness (inside water) to its full weight (in air) over the distance of its Center of Mass shift.
The Center of Mass moves from depth $r$ (just touching the surface from below) to height $r$ (just touching the surface from above). However, the buoyant force decreases linearly as it exits.
Alternative Energy Approach:
Since the sphere “replaces” water, lifting the sphere out is equivalent to lifting a sphere of water from the bulk reservoir to the position just above the surface. The mass of this water is $m$ (since densities are equal). The Center of Mass of this volume is raised by a height equal to the radius $r$.
$$W = \Delta PE = m g \times (\text{Height of COM above surface})$$ $$W = m g r$$