question_answer

The pressure at a depth of 13.6 cm in a water beaker is same as the pressure at a depth of 1 cm in another beaker containing an unknown liquid. Identify the unknown liquid.          

Answer:

                           

Case - 1 Water	Case - 2 Un known liquid
\[{{h}_{1}}=13.6\text{ }cm\]	\[{{h}_{2}}=1\text{ }cm\]
\[{{P}_{1}}=P\]	\[{{P}_{2}}=P\]
\[{{d}_{1}}=1g\text{ }c{{m}^{3}}\]	\[{{d}_{2}}=?\]
\[{{g}_{1}}=g\]	\[{{g}_{2}}=g\]

We know,                           \[P=hdg\] Both the liquids are exerting same pressures. So, \[{{P}_{1}}={{P}_{2}}\] \[{{h}_{1}}{{d}_{1}}{{g}_{1}}={{h}_{2}}{{d}_{2}}{{g}_{2}}\] \[\Rightarrow \] \[{{h}_{1}}{{d}_{1}}={{h}_{2}}{{d}_{2}}\]                      (At a given place\[{{g}_{1}}={{g}_{2}}\]) \[\Rightarrow \]\[{{d}_{2}}=\frac{{{h}_{1}}{{d}_{1}}}{{{h}_{2}}}=\frac{13.6\times 1}{1}=13.6\,g\,c{{m}^{-3}}\] \[13.6\text{ }g\text{ }c{{m}^{3}}\]is the density of mercury, So, the unknown liquid is mercury.