Question

Given that the density of water is 1gm/cm3, the pressure that can support a column of water 100cm high can also support a column of 80cm high of a given sugar solution. What is the density of the sugar solution

Answer 1

Answer:

$1250\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$

Explanation:

We know that the pressure is calculated by,

$P = \rho gh$

where h is the height of the column and $\rho$ is the density of the fluid.

For Water:

Pressure is,

$\begin{align*}{P_1} &= {\rho _w}g{h_1}\\ &= \left( {1000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)g\left( {1\;{\rm{m}}} \right)\\ &= \left( {1000\;{\rm{kg/}}{{\rm{m}}^2}} \right)g\end{align*}$

For Sugar solution:

Pressure is,

$\begin{align}{P_2} &= {\rho _2}g{h_2}\\ &= {\rho _2}g\left( {0.8\;{\rm{m}}} \right)\end{array}$

As the pressure can support both columns of water and sugar solution,

$\begin{align}{P_1} &= {P_2}\\\left( {1000\;{\rm{kg/}}{{\rm{m}}^{\rm{2}}}} \right)g &= {\rho _2}g\left( {0.8\;{\rm{m}}} \right)\\{\rho _2} &= \dfrac{{1000\;{\rm{kg/}}{{\rm{m}}^{\rm{2}}}}}{{0.8\;{\rm{m}}}}\\{\rho _2} &= 1250\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\end{array}$

The density of sugar solution is:$1250\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$