A wooden ball fell into a high cylindrical vessel. The diameter of the vessel and the diameter of the ball is 5 cm and 4 cm respectively. Can you get the ball without touching the vessel?

###### Solution. Posted on Friday, June 12, 2015.

It is commonly known fact that wood is floating in the water. We can use this fact to get the ball without touching the vessel. If we pour the water into the vessel then the ball will float upward. Of course, if the buoyant force is greater than the weight of the wooden ball.

Let’s calculate it.

The radius of the wooden ball is 2 cm = 0.02 m. The volume of the ball is given by the formula

V = 4/3*π*R^3. So, 4/3*π*0.02^3 = 3.35*10^-5 m^3.

The density of the wood is 630 kg/m^3 then based on the following formula

W = ρ*V*g we could get the weight of the ball.

W = ρ*V*g = 630*3.35*10^-5*9.81 = 0.207 N

Now, we can calculate the buoyant (Archimedes) force.

F = ρ*V*g , where ρ – the density of the water (1000 kg/m^3) and V – is the volume of the displaced body of water.

F = ρ*V*g = 1000*3.35*10^-5*9.81 = 0.329 N

Finally, F > W or the buoyant force is greater than the weight of the wooden ball.

The ball will float upward and we could pick up it from the top of the vessel.

Keep adding water in the cylinder until ball pops out …as it being made of wood it will float.

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Mathematically, it is possible as the surface area of the cylinder’s base is 6.25pi cm^2 (pi x 5/2^2) and the ball’s widest spread of area is 4pi cm^2 (pi x 4/2^2) which is at its equator. So, since the ball’s maximum surface area at any level is smaller than the cylinder’s base area, it is possible for the ball to be taken out without touching the vessel.

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