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 0.4 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/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)

0.05m

Explanation:

Density of water = ρ(w) = 1000 kg/ m³ ;  

Density of Mercury = ρ(m) = 13628.95 kg/ m³  

Total pressure at bottom of cylinder=1.1atm

Therefore, pressure due to water and mercury =1.1-1 =atm

0.1atm=10130pa

The pressure at the bottom is given by,

ρ(w) x g[0.4 - d] + ρ(m) x g x d  = 10130

1000 x 9.8[0.4 - d] + 13628.95 x 9.8 d = 10130

3924 - 9810d + 133416d= 10130

123606d= 6206

d= 6202/123606

d= 0.05m

Depth of mercury alone =  d = 0.05m

as technology advances, new experiments often expose problems in accepted theories. c

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-lavira: )

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.