Question

A sphere of mass $m$ and radius $r$ rests at the bottom of a large reservoir of water as shown in figure-I. Depth of the reservoir is $h$. Density of material of the sphere is the same as that of water. Now the sphere is slowly pulled completely out of water as shown in figure-II.

Which of the following is a correct expression for work done by the agent pulling the sphere?

• A. $mgr$
• B. $0.5mgr$
• C. $mg(r+h)$
• D. $mg(0.5r+h)$

Answer 1

Answer: (A)

mgr

Answer 2

The work done by the agent pulling the sphere slowly is equal to the change in the total potential energy of the sphere-water system.

$W_{agent} = \Delta U_{system} = \Delta U_{sphere} + \Delta U_{water}$

1. Change in potential energy of the sphere ($\Delta U_{sphere}$):

Let the bottom of the reservoir be the reference level for potential energy ($y=0$).

Initial position of the sphere's center of mass: $y_i = r$.

Final position of the sphere's center of mass (completely out, resting on the surface): $y_f = h+r$.

$\Delta U_{sphere} = mg(y_f - y_i) = mg((h+r) - r) = mgh$.

2. Change in potential energy of the water ($\Delta U_{water}$):

The density of the sphere is the same as that of water ($\rho_{sphere} = \rho_{water}$).

The volume of the sphere is $V = \frac{4}{3}\pi r^3$.

The mass of the sphere is $m = \rho_{sphere} V$.

When the sphere is pulled out of the water, the volume $V$ of water that was displaced by the sphere drops down to fill the space previously occupied by the sphere. Since $\rho_{water} = \rho_{sphere}$, the mass of this water is also $m$.

Initially, this mass of water $m$ was effectively displaced upwards. Its center of mass was at the water surface level, $y=h$.

Finally, this mass of water $m$ fills the space previously occupied by the sphere at the bottom. The center of mass of this volume of water is now at $y=r$.

Therefore, the change in potential energy of this water is:

$\Delta U_{water} = mg(y_{water,final} - y_{water,initial}) = mg(r - h)$.

3. Total Work Done by the Agent:

$W_{agent} = \Delta U_{sphere} + \Delta U_{water}$

$W_{agent} = mgh + mg(r - h)$

$W_{agent} = mgh + mgr - mgh$

$W_{agent} = mgr$

The work done by the agent is $mgr$.