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Two immiscible liquids of densities rho ...

Two immiscible liquids of densities `rho` and `2rho` and thickness (heights) `h` and `2h`, respectively, push a vertical plate. Find the total side thrust given by the liquids on the vertical plate.

Text Solution

Verified by Experts

The correct Answer is:
`F=13/2rhogh^(2)b`

Let the hydrostatic forces acting on the rectangular pathes of areas `A_(1)` and `A_(2)` of the gate be `F_(1)` and `F_(2)`. Then the total hydrostatic force on the gate is
`F=F_(1)+F_(2)`
where `F_(1)=P_(av_(1))A_(1)` and `F_(2)=P_(av_(2)) A_(2). P_(av_(1))` and `P_(av_(2))` and `P_(av_(2))` are the average pressures on areas `A_(1)` and `A_(2)`, respectively.
`F=P_(av_(1))A_(1)+P_(av_(2))A_(2)`

Substituting `P_(av1)=(P_(A)+P_(B))//2, A_(1)=bh_(1)=bh, P_(av2)=(P_(B)+P_(C)).2,A_(2)=bh_(2)=b(2h),` we have
`F=((P_(A)+P_(B)))/2bh+((P_(B)+P_(C)))/2 2bh`
Substituting `P_(A)=0, P_(B)=rhogh` and `P_(C)=(rho_(1)h_(1)+rho_(2)h_(2))g=[rhoh+(2rho)(2h)]=5rhogh,` we have
`F=3/2rhogh^(2)b`
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Knowledge Check

  • A vertical pole of length l , density rho , area of cross section A , floats in two immiscible liquids of densities rho_(1) and rho_(2) . In equilibrium position the bottom end is at the interface of the liquids. When the cylinder is displaced vertically, the time period of oscillation is..........

    A
    `pisqrt((rhol)/((rho_(1)+rho_(2))g))`
    B
    `pi sqrt((rhol)/(g)) ((1)/(sqrt(rho_(1)))+(1)/(sqrt(rho_(2))))`
    C
    `pi sqrt(((rho_(1)+rho_(2))l)/(rhog))`
    D
    `pi sqrt((l)/(rhog)) (sqrt(rho_(1))+sqrt(rho_(2)))`
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    A
    `1/(2sqrt(2))`
    B
    `1/2`
    C
    `1/4`
    D
    `1/(sqrt(2))`
  • In a cylindrical vessel containing liquid of density rho , there are two holes in the side walls at heights of h_1 and h_2 respectively such that the range of efflux at the bottom of the vessel is same. The height of a hole, for which the range of efflux would be maximum, will be –

    A
    `h_(2)-H_(1)`
    B
    `H_(2)+H_(2)`
    C
    `*h_(2)-h_(2))/2`
    D
    `(h_(2)+h_(1))/2`
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