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

snell's law

refractve index = sin incidence / sin refraction

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation