Đáp án:

a. \(m = \dfrac{{{S_1}{m_2} - {m_1}{S_2}}}{{{S_2}}}\)

b. \(h' = \dfrac{{{S_1} + {S_2}}}{{{S_2}}}h\)

Giải thích các bước giải:

Ta có:

\[h = \dfrac{{10}}{{{d_n}}}\left( {\dfrac{{{m_2}}}{{{S_2}}} - \dfrac{{{m_1}}}{{{S_1}}}} \right)\]

a. Để hai bên nước cân bằng nhau thì:

\[\begin{array}{l}
\dfrac{{{F_1}}}{{{S_1}}} = \dfrac{{{F_2}}}{{{S_2}}}\\
 \Leftrightarrow \dfrac{{10\left( {{m_1} + m} \right)}}{{{S_1}}} = \dfrac{{10{m_2}}}{{{S_2}}}\\
 \Leftrightarrow m = \dfrac{{{S_1}{m_2} - {m_1}{S_2}}}{{{S_2}}}
\end{array}\]

b. Khi đặt quả cân m vào bên đối diện thì lúc này:

\[\begin{array}{l}
\Delta p = {p_2} - {p_1}\\
 \Leftrightarrow {d_n}h' = \dfrac{{10\left( {m + {m_2}} \right)}}{{{S_2}}} - \dfrac{{10{m_1}}}{{{S_1}}}\\
 \Leftrightarrow h' = \dfrac{{10}}{{{d_n}}}\left( {\dfrac{{\dfrac{{{S_1}{m_2} - {m_1}{S_2}}}{{{S_2}}} + {m_2}}}{{{S_2}}} - \dfrac{{{m_1}}}{{{S_1}}}} \right)\\
 \Leftrightarrow h' = \dfrac{{10}}{{{d_n}}}.\dfrac{{{S_1}{m_2} - {S_2}{m_1}}}{{{S_2}^2}} + h\\
 \Leftrightarrow h' = \dfrac{{{S_1}}}{{{S_2}}}.\dfrac{{10}}{{{d_n}}}\left( {\dfrac{{{m_2}}}{{{S_2}}} - \dfrac{{{m_1}}}{{{S_1}}}} \right) + h\\
 \Leftrightarrow h' = \dfrac{{{S_1} + {S_2}}}{{{S_2}}}h
\end{array}\]