A sphere with a surface area of $195.8 \mathrm{dm}^{2}$ is floating on $+4 \mathrm{C}$ water, submerged to a depth of $1.2 \mathrm{dm}$. What is the area of the submerged part of the surface, and what is the weight of the sphere? What is the central angle of the spherical sector corresponding to the submerged surface?
The radius of the sphere

$$
r=\sqrt{\frac{195.8}{4 \pi}}=3.948 \mathrm{dm}
$$

The submerged surface area:

$$
f=2 r \pi m=29.75 \mathrm{dm}^{2}
$$

The weight of the sphere is equal to the weight of the displaced water, that is:

$$
Q=\pi m^{2}\left(r-\frac{m}{3}\right)=16.04 \mathrm{~kg}
$$

Denoting the apex angle of the spherical sector by $\alpha$:

$$
\cos \frac{\alpha}{2}=\frac{r-m}{r}=\frac{2.748}{3.948}
$$

and thus

$$
\alpha=91^{\circ} 47^{\prime} 18^{\prime \prime}
$$

(József Sasvári, Pécs.)

The problem was also solved by: K. Czank, J. Filkorn, E. Freibauer, M. Glass, T. Kerekes, A. Kiss, B. Kohn, B. Krausz, Gy. Krisztián, Gy. Lukhaub, J. Neumann, R. Obláth, J. Porkoláb, P. Rozlosnik, J. Sasvári, Ö. Spitzer.