 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        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

```							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

```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »  ### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions