Two communicating cylindrical tubes contain mercury. The diameter of one vessel is four times large than the diameter of the outer. A column of water of height `70 cm` is poured into the narrow vessel. How much will the mercury level rise in the other vessel and how much will it sink in the narrow one? How much will the mercury level rise in the narrow vessel, if a column of water of the same height is poured into the broad vessel?

Explanation

Let `x=`rise of `Hg` level in the broad vessel, `y=`drop of Hg level in the narrow vessel. Then,
`xA_(1)=yA_(2)implies16x=y`
(because volume displaced will be the same)
Equating pressures at `A` and `B` on the same horizontal plane, we get
`70xx1xxg=(x+y)xx136xxgimpliesx+y=70/13.6`
Solving the equations, we get `x=0.3 cm y=4.8 cm`. If `70 cm` water column is pured into the broad vessel, then following the same pattern, we get the equations
`x'=16y'` and `(x'+y')xx13.6=70xx1`
`impliesx'=4.8cm` and `y'=0.3cm`