Giải bài 3 trang 266 – Bài 54 - SGK môn Vật lý lớp 10 Nâng cao

$\begin{align} & {{h}_{1}}=80\text{ mm; }\rho =\text{ 1}{{\text{0}}^{3}}\text{ }\left( kg/{{m}^{3}} \right);\text{ }{{\sigma }_{1}}=0,072\text{ }\left( N/m \right) \\ & {{h}_{2}}=?;\text{ }{{\rho }_{2}}=790\ \left( kg/{{m}^{3}} \right);\ {{\sigma }_{2}}=0,022\text{ }\left( N/m \right) \\ & {{h}_{1}}=\dfrac{4}{gd}.\dfrac{{{\sigma }_{1}}}{{{\rho }_{1}}};\text{ }{{\text{h}}_{2}}=\dfrac{4}{gd}.\dfrac{{{\sigma }_{2}}}{{{\rho }_{2}}} \\ & \Rightarrow \dfrac{{{h}_{2}}}{{{h}_{1}}}=\dfrac{{{\sigma }_{2}}{{\rho }_{1}}}{{{\sigma }_{1}}{{\rho }_{2}}}\Rightarrow {{h}_{2}}=\dfrac{{{h}_{1}}.{{\sigma }_{2}}.{{\rho }_{1}}}{{{\sigma }_{1}}.{{\rho }_{2}}}=\dfrac{80.0,{{022.10}^{3}}}{0,072.790}=30,9\text{ }\left( mm \right). \\ \end{align}$