A `U`-tube of base length `l` filled with the same volume of two liquids of densities `rho` and `2rho` is moving with an acceleration `'a'` on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height `h` is given by

Options

A. `a/(2g)l`
B. `(3a)/(2g)l`
C. `a/gl`
D. `(2a)/(3g)l`

Answer

B

Explanation

For the given situation liquid of density `2rho` should be behind that of `rho`
From the right limb `P_(A)=P_(atm)+rhogh`
`P_(B)=P_(A)+rhoa1/2=P_("min")+rhogh+rhoal/2`
`P_(C)=P_(B)+(2rho)al/2=P_(atm) +rhogh+3/2rhoal`.........i
But from the lift limb `P_(C)=P_(atm)+(2rho)gh`...........ii
From eqn i and ii
`P_(atm)+rhogh+3/2rhoal=P_(atm)=2rhogh`
`implies h=(3al)/(2g)`