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Grade: 11
        
a rectangular vessel is filled with water and oil in equal proportion,the oil beingtwice lighter than water. Show that force on each side of the wall of the vessel will be reducedby one fifth if the vessel is filled only with oil.
 
2 years ago

Answers : (1)

Arun
23769 Points
							
Dear Dishank
 

In this question there is container having water and oil of equal quantity.

we have to calculate the force on the side wall of the container.

Let width of side wall = b

height of side wall = 2h

ρ(water) = 2ρ(oil)

h = height of water in the container

As we know that the volume of oil and water are same. Then

h = height of water in the container.

The height will be same for both the fluids.

area of wall covered by oil = bh

area of wall covered by water = bh

Now As the oil is lighter then it will lie above water.

Now first of all i would like to tell you the force exerted by the fluid on a vertical plate.

F = ρg(h1)A

where A = Area of the vertical plate on which force is acting

h1 = centroid of the area from the free surface of the liquid.

Let width of the side wall of container = b

Area covered by the oil on the side wall = bh

Now there will be two forces exerted on the side wall. These are -

F1 = force due to oil

F2 = force due to water

F1 = ρ(oil) x g x h1 x A = ρ(oil) x g x (h/2) x bh = ρ(oil)ghb²/2

Now we have to calculate the force due to water on the wall.

F2 = ρ(water) x g x (h1) x bh

Now we have to calculate the h1 in the formula

Always make sure that when you are calculating h1 for a volume of fluid then the fluid above it should be same. For example here in this question there is oil present above water so we cannot calculate h1 for water directly. First we will have to remove oil present above it and should fill equivalent water having same pressure as that of oil.

Now pressure due to oil = ρ(oil) x g x h

Pressure due to water = ρ(water) x g x h2

where h2 is the equivalent height if water is used in place of oil.

ρ(oil) x g x h = ρ(water) x g x h2

ρ(oil) x g x h = 2 x ρ(oil) x h2

h2 = h/2

Now we replace oil with equivalent water having height h/2.

Now earlier i told you that h1 = height of the centroid of area from the free surface of water

so, h1 = h/2 + h2 = h/2 + h/2 = h

we get

F2 = ρ(water) x g x h1 x bh = 2ρ(oil)gbh²

Now total force = F1 + F2 = ρ(oil)gbh²/2 + 2ρ(oil)gbh² = 5ρ(oil)gbh²/2

Now we are calculating force if the container is fully filled with oil

we get

F´ = ρ(oil) x g x h1 x A

Now here Area A = b(2h) = 2bh

h1 = centroid of area (2bh) from the free surface = h

F´ = ρ(oil) x g x h x 2bh = 2ρ(oil)gbh²

Reduction in force = F - F´ = 5ρ(oil)gbh²/2 - 2ρ(oil)gbh² =ρ(oil)gbh²/2

Ratio = (F - F´)/F = 1/5

 

Regards

Arun (askIITians forum expert)

2 years ago
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