Calculate the height of a column of liquid glycerol (d = 1.26 g/cm³),in meters,required to exert the same pressure as 4.91 m of water. A)4.91 m B)3.90 m C)6.19 m D)3.65 m E)5.16 m

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Quizplus · Accepted Answer

The pressure exerted by a column of liquid is given by $P = h \cdot \rho \cdot g$, where $P$ is the pressure, $h$ is the height of the liquid column, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity. To exert the same pressure, the product of the height and the density of the two liquids must be equal. Therefore, $h_{glycerol} \cdot \rho_{glycerol} = h_{water} \cdot \rho_{water}$. Given that $\rho_{water} = 1.00 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3$ and $\rho_{glycerol} = 1.26 \, \text{g/cm}^3 = 1260 \, \text{kg/m}^3$, and $h_{water} = 4.91 \, \text{m}$, we can solve for $h_{glycerol}$ as follows: $h_{glycerol} = \frac{h_{water} \cdot \rho_{water}}{\rho_{glycerol}} = \frac{4.91 \cdot 1000}{1260} = 3.90 \, \text{m}$.

Calculate the Height of a Column of Liquid Glycerol (D