# Sunday Times Teaser 3016 – Eureka

### by Andrew Skidmore

#### Published Sunday July 12 2020 (link)

Archimedes’ Principle states that the upward force on a floating body equals the weight of water displaced by the body. In testing this, Liam floated a toy boat in a cylindrical drum of water. The boat had vertical sides and took up one tenth of the surface area of the water. He also had some identical ball bearings (enough to fit snugly across the diameter of the drum), which were made of a metal whose density is a single-figure number times that of water.

Liam loaded the boat with the ball bearings and noted the water level on the side of the drum. When he took the ball bearings out of the boat and dropped them in the water, the water level changed by an amount equal to the radius of a ball bearing.

How many ball bearings did Liam have?

Let there be $$N$$ ball bearings of radius $$r$$ and density $$\rho$$.   This means that the total volume of the ball bearings is $$(4/3)\pi r^3 N$$ and the volume of the same weight of water is $$\rho$$ times this.  The drum radius is $$Nr$$, giving the volume of water in a drum height of $$h$$ as $$\pi r^2 N^2 h$$.

When the ball bearings are dropped into the water the change in its height $$h_1$$ is given by:$\pi r^2 N^2 h_1 = \frac{4}{3}\pi r^3N$ which gives $$h_1=4r/3N$$.

When the ball bearings are placed in the boat, the boat will sink until the increased pressure is sufficient to support their weight. If it sinks by a length $$h$$ we can equate the increased water pressure to the pressure on the bottom of the boat created by the ball bearings:$h^{‘} g=\frac{(4/3)\pi r^3N\rho g}{\pi r^2 N^2/10}$where $$g$$ is the acceleration due to gravity. This gives $$h^{‘}=(40r\rho/3N)$$. When the bearings are taken out the boat rises and the displaced water spreads across the whole drum and the height drops to $$h^{”}$$ such that $$h^{”} A=(9/10)h^{‘}A$$ where $$A$$ is the area of the drum’s base. So the height change when the ball bearings are loaded is:$h^{‘} – h^{”} = \frac{4r\rho}{3N}$Equating the difference in these two heights to the ball radius now gives: $3N=4(\rho-1)$Since $$\rho$$ is a single digit number this show immediately that $$\rho$$ is either 4 or 7 and $$n$$ is either 4 or 8.

It appears that the teaser author wrongly assumed that the displaced water would end up in the annular ring of water round the boat so that:$(9/10)\pi r^2Nh=(4/3)\pi r^3 N \rho$This gives the change in the height of the water as $$(40r\rho/27N)$$ which leads to a different equation:$27N=4(10\rho-9)$with the single solution of $$\rho=9$$ leading to $$n=12$$ as the intended solution. I also made the same mistake in my initial solution.

There are two states to consider:

(2) empty boat; floating; balls on bottom of drum

Let there be m balls of radius 1 and relative density n.
Let U be the volume displaced by the empty boat and V be the volume of water.
The radius of the drum is m.

The volumes below the waterline are:

(1) U+V+4πmn/3
(2) U+V+4πm/3

The difference is πm^2 = 4πm(n-1)/3

So 3m = 4(n-1) giving (m,n) = (4,4) or (8,7).

The incorrect “officially correct” third solution m = 12, would be correct for relative density 10.

As long as the boat is able to contain the balls the surface area of the water taken up by the boat is irrelevant and so is its shape.

I was amused by speculation about the size. shape and loading pattern of the boat. The only unit referred to is “size of a ball bearing”. I found a list of available ball bearing diameters ranging from 0.4mm to 108mm.

Although it seems to be a technical term in electronics and botany, the phrase “annular ring” is a tautology.