A vessel of height `H` and length `L` contains a liquid of density `rho` upto height `H//2` The vessel start accelerating horizontally with acceleration `a` towards right. If A is the point at the surface of the liquid at right . If A is the point at the surface of the liquid at right end while the vessel is accelerating and B is the point at bottom of the vessel on the other end, the difference of pressures at B and A will be

Options

A. `(rho)/(2)(gH+aL)`
B. `(rho)/(2)(gH-aL)`
C. `2rho(gH-aL)`
D. `(3rho)/(2)(gH+aL)`

Answer

A

Explanation

Let force `(DeltaP)` A acts on the surface of the liquid while vessel is accelerating.
`(DeltaP.A)costheta=mg,(DeltaP.A)sintheta=ma`
`tantheta=(a)/(g)` and `y=(L)/(2)tantheta=(aL)/(2g)`
`h_(1)=(H)/(2)-(aL)/(2g)`
Considering a fluid element at distance `x` from left side of the vessel then
`(p+dp)A^(')-pA^(')=(A^(')dxrho)a`
`int_(A)^(C)dp=int_(A)^(C)rhoadx,p_(C)-p_(A)=arhoL`
If points A and C are at same level
`p_(B)-p_(C)=rhogh_(1)=rhog[(H)/(2)-(aL)/(2g)]`
`p_(B)-(rhoghL+p_(A))=rhog[(H)/(2)-(aL)/(2g)]`
`p_(B)-p_(A)=(rho)/(2)(gH+aL)`