A solid, homogeneous cylindrical body floats in water such that its axis is horizontal and the depth of submersion is equal to half the radius of the base. What is the specific weight of the material?
Let the specific gravity of the body be $F$, its volume $K$, the specific gravity of the fluid $f$, and the volume of the fluid displaced by the body $k$. According to Archimedes' principle,

$$
K F = k f
$$

In this case, $f$ can be taken as 1, and thus the specific gravity we are looking for is

$$
F = \frac{k}{K}
$$

Let the area of the base circle of the cylinder be denoted by $T$, and its height by $m$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2b8e0bb2458dd4aedb5fg-1.jpg?height=368&width=385&top_left_y=510&top_left_x=859)

The displaced water can be considered as a right cylinder whose base is a circular segment with area $t$ and height $m$. Therefore,

$$
F = \frac{k}{K} = \frac{t m}{T m} = \frac{t}{T}
$$

If the radius of the base circle is denoted by $r$, then $T = r^2 \pi$. The central angle corresponding to the arc of the circular segment is denoted by $\alpha$. According to the problem (1st figure),

$$
\cos \frac{\alpha}{2} = \frac{\bar{r}}{r} = \frac{1}{2}, \quad \text{or} \quad \frac{\alpha}{2} = 60^\circ \quad \text{and thus} \quad \alpha = 120^\circ
$$

Accordingly,

$$
t = \frac{r^2 \pi}{3} - \frac{r}{2} \cdot \frac{r}{2} \sqrt{3} = r^2 \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right)
$$

Thus, the specific gravity we are looking for is

$$
F = \frac{t}{T} = \frac{r^2 \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right)}{r^2 \pi} = \frac{1}{3} - \frac{\sqrt{3}}{4 \pi} \approx 0.1955 \, \text{g} \, \text{cm}^{-3}
$$

Béla Rozsondai (Sopron, Széchenyi g. IV. o. t.)