Question

Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water. If the cylinder is 0.9 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m3, and the ambient atmospheric pressure to be 1.013e5 Pa)

Answer 1

Answer:0.0106 m

Explanation:

Given

Absolute Pressure at the bottom is $P=1.1 atm$

Height of cylinder is $h=0.9 m$

Density of mercury $\rho _{hg}=13.6\times 10^{3} kg/m^3$

Pressure Difference is given by

$\Delta P=\rho g\cdot h_0$

Pressure at bottom=Pressure at Top+Pressure due to mercury and water

$1.1\times 1.0132\times 10^5=1.01325\times 10^5+\rho _w\times 9.8\times (0.9-d)+\rho _{hg}\times 9.8\times d$

converting Pressure into KPa

$111.45=101.325+9.8\cdot (0.9-d)+13.6\cdot 9.8\cdot d$

$\frac{10.13}{9.8}=0.9 -d +13.6 d$

$d=0.0106 m$