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A large container contains a liquid of d...

A large container contains a liquid of density p, mass m, and placed on a smooth horizontal floor. The mass of container is negligible and cross sectional area is A. Now at the bottom of the container, there is small hole of cross section area A/200 in the side wall through which the liquid starts flowing horizontally at t=0. What will be the velocity of efflux when half of the liquid is drained out. Also, find the acceleration of the liquid flowing out.

Text Solution

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h=height of liquid in the container
`h=("Volume")/("Area")=(m//p)/(A)=(m)/(pA)`
Here, m=mass of liquid, p=density of liquid
Initially the container is completely filled with liquid
Let V=velocity of efflux =`sqrt(2gh)`
`=sqrt((2gm)/(pA))`
Let the liquid starts flowing out from the hole.
Let m.= mass of liquid flowing out through hole in 1 second `m.=((A)/(200)) pv=(pvA)/(200)`
`triangleP` =momentum contained by the liquid/sec.
`triangle P=m.v=(pvA)/(200), v=(pv^(2) A)/(200)`
As by Newton.s second law of motion
Rate of change of momentum/sec of container =Force acting on the container `(triangleP)/(trianglet)=F`
For 1 sec, `F=(triangleP)/(1)=(pA v^(2))/(200) ........(i)` br> When 50% of the liquid is drained out then `h.=h/2`
Velocity of efflux `=sqrt(2gh.)=sqrt((2gh)/2)`
`V=sqrt((2gm)/(2pA))=sqrt((gm)/(pA))..........(ii)`
Acceleration `=F/m=(pA v^(2))/(200 m)` (Using (i))
Using (ii) Acceleration `=(pA gm)/(pA 200 m)=(g)/(200)`
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