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Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

The various faces on which the cuboid can be turned are 2*3(6) or 3*4(12) since the cuboid is
already on its third face which is 2*4(8). The smallest face towards which the tank is moved
has an area of 6 meter^2.

Therefore, the depth of water is meter in the cube can be measured as \(\frac{4}{3}\) meter(Option C)
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pushpitkc


Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

The various faces on which the cuboid can be turned are 2*3(6) or 3*4(12) since the cuboid is
already on its third face which is 2*4(8). The smallest face towards which the tank is moved
has an area of 6 meter^2.

Therefore, the depth of water is meter in the cube can be measured as \(\frac{4}{3}\) meter(Option C)


pushpitkc I omitted this step / couldnt see it ---> Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

but i did calculate the are of large rectangle which is 8

and area of smallest rectangle which is 6

so al we have to do is to write ratio the areas of above mentioned rectangles ? if So

my question IS why you wrote 4/3 and not 3/4, how should i figure out this - to write 4/3 or 3/4 ?

i hope you can take time to reply to my post :grin:

thank you :)
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Hey dave13

Once we have established the volume of water contained in the rectangular tank is 8 m^3
and the smallest face on which the rectangular tank is turned is 2*3 or 6 m^2, in order to
understand how high the water is the calculations will be as follows

6m^2 * height of water(in m) = 8m^3 (volume of water we need to fit in the vessel)

Now the height in meters is \(\frac{8}{6} = \frac{4}{3}\)

Hope that helps!
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dave13
pushpitkc


Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

The various faces on which the cuboid can be turned are 2*3(6) or 3*4(12) since the cuboid is
already on its third face which is 2*4(8). The smallest face towards which the tank is moved
has an area of 6 meter^2.

Therefore, the depth of water is meter in the cube can be measured as \(\frac{4}{3}\) meter(Option C)


pushpitkc I omitted this step / couldnt see it ---> Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

but i did calculate the are of large rectangle which is 8

and area of smallest rectangle which is 6

so al we have to do is to write ratio the areas of above mentioned rectangles ? if So

my question IS why you wrote 4/3 and not 3/4, how should i figure out this - to write 4/3 or 3/4 ?

i hope you can take time to reply to my post :grin:

thank you :)

Hi dave13,

I think you are confused with the application of Area and Volume.

In the above problem, the quantity(volume ) of water is going to be constant. Hence we equate the volume.

In the Original Diagram, we have the following dimensions:

Height = 3 cm but the height of the water is only 1 cm, hence consider height as 1 cm. Length = 2 cm and Width = 4 cm.

Volume = 2*4*1 = 8 \(cm^3\) ---- (1)

After rotation, we have the following diagram. Please note that the water level is not depicted in the diagram.

Attachment:
rotated tank1.png
rotated tank1.png [ 17.19 KiB | Viewed 6938 times ]

After rotation we have the following dimensions:

Height = 4 cm but the height of water level is unknown, hence assume as \(h\) cm. Length = 3 cm, and Width = 2 cm.

The volume of water =\(h*3*2 = 6h\) --- (2)

By equating (1) and (2) we have \(h = \frac{4}{3}\)cm.

Hope this helps.

Please go through the following article:

https://study.com/academy/lesson/differ ... olume.html

Thank you.
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ganand and pushpitkc thanks a lot :)
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