A vertical `U`-tube has two liquids `1` and `2`. The heights of liquids columns in both the limbs are `h` and `2h`, as shown in the figure. If the density of the liquid `1` is `2rho`.

a. Find the density of liquid `2`.
b. If we accelerate the tube towards right till the heights of liquid columns will be the same, find the acceleration of the tube.

Answer

`3/5g`

Explanation

First of all we take two points `1` and `2` at the bottom of the left and right vertical limbs. The pressure at `1` is
`P_(1)=P_(0)+(2rho)gh`

The pressure at `2` is `P_(2)=P_(0)+rho'g(2h)`
where `rho'=`density of the other liquid
since `P_(1)=P_(2)` for stationary liquid, we have
`P_(0)+2rhogh=P_(0)+rho'g(2h)`
This gives `rho'rho`.
b. If the levels in both the limbs eqal, conserving the volume of the total (two) liquids, the height of each liquid column will be equal to `3h//2`. The length of liqids in the horizontal limbs are `h//2` and `3h//2`, respectively.
The pressure at interface `3` of the liquid is
`P_(3)=P_(1)-(2rho)a(h/2)`
Where `P_(2)=P_(0)+(2rho)a((3h)/2)`
This gives `P_(3)=P_(0)+(2rho)g((3h)/2)-(2rho)a(h/2)`
`=P_(0)+3rhogh-rhoah`.........i
`P_(3)` can also be give as `P_(3)=P_(2)+rhoa((3h)/2)` where `P_(2)=P_(0)+rhoa`
`((3h)/2)`. This gives
`P_(3)=P_(0)+3/2rhogh+3/2rhoah`..........ii
Eliminating `P_(3)` from eqn i and ii we have
`a=3/5g`