A $16 \mathrm{~cm}$ diameter solid oak ball floats in water such that $307.2 \mathrm{~cm}^{2}$ of the sphere's surface remains dry. What is the specific gravity of the oak?

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The translation is provided as requested, maintaining the original formatting and structure.
Let's denote the volume of the sphere by $K$, the volume of the submerged spherical segment by $k$, and the specific gravity we are looking for by $x$. According to Archimedes' principle,

$$
K \cdot x = k \cdot 1
$$

from which

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

The height of the spherical cap that remains dry is denoted by $m_{1}$,

$$
2 r \pi m_{1} = 307.2
$$

from which

$$
m_{1} = \frac{307.2}{16 \pi} \sim 6.113 \text{ cm}
$$

and thus the height of the submerged spherical segment is

$$
m_{2} = 16 - m_{1} = 9.887 \text{ cm}
$$

Therefore,

$$
k = \frac{m_{2}^{2} \pi}{3} \left(3 r - m_{2}\right) = \frac{9.887^{2} \pi}{3} \cdot 14.113
$$

On the other hand,

$$
K = \frac{4 \cdot 8^{3} \pi}{3}
$$

and thus

$$
\begin{aligned}
& x = \frac{k}{K} = \frac{9.887^{2} \cdot 14.113}{4 \cdot 8^{3}} = \frac{9.887^{2} \cdot 14.113}{2048} \\
& 2 \lg 9.887 = 1.9902 \\
& \lg 14.11 = 1.1495 \\
& \Sigma = 4.1397 - 1 \\
& -\lg 2048 = 3.3114 \\
& \lg x = 0.8283 - 1 \\
& x = 0.6734
\end{aligned}
$$

Therefore, the specific gravity of the oak is $0.6734 \frac{\text{grs}}{\text{cm}^{3}}$.

Argay Gyula (Balassagyarmat, Balassa g. IV. o. t)

Note: Most of the students unnecessarily calculated intermediate results numerically.