An equal-armed vessel, where the cross-section of the arms is equal, has been filled with mercury so that it rises to a sufficient height in both arms. If now we pour a $v$ volume, $s$ specific gravity liquid into the mercury in one of the arms, at what vertical height do the surfaces of the liquids stand in the two arms relative to each other?

(Teacher Szontágh Gusztáv from the main real school in Brassó.)
After the balance is restored, the weight of the mercury column, which stands on the horizontal plane separating the liquid column from the mercury level, will be equal to the weight of the liquid column. In symbols,

$$
v s=v^{\prime} S^{\prime}
$$

But

$$
\begin{gathered}
h: h^{\prime}=v: v^{\prime} \\
h-h^{\prime}: h=v-v^{\prime}: v \\
h-h^{\prime}=\frac{h}{v}\left(v-v^{\prime}\right)=\frac{h}{v}\left(v-\frac{V S}{S^{\prime}}\right) \\
h-h^{\prime}=\frac{h}{S^{\prime}}\left(S^{\prime}-S\right)
\end{gathered}
$$

where $h$ and $h^{\prime}$ represent the heights of the liquid and mercury columns, respectively.

Sztrapkovits István, S.-A.-Ujhely, Kugel Sándor, Losoncz, Kutlik Igor, Pozsony.