Question

A sealed rectangular container 10 inches by 12 inches by 6 inches is sitting on its largest face. If it is filled with water up to a level 2 inches from the top, how many inches from the bottom will the water level reach if the container is placed on its smallest face?

Answer 1

Answer:

8 inches

Step-by-step explanation:

Given: Dimension of rectangular container= $(10\times12\times 6)\ inches$

           Water is filled up to a level 2 inches from the top.

As given that container is sitting on its largest face, then it is standing on $(10\times 12)\ inches$

Which means the height of container is 6 inches and it ie filled with water up to 2 inches from the top.

∴ Height of water inside the container= $6-2= 4 \ inches$

Now, finding the volume of water inside.

we know, Volume= $length\times width\times height$

Volume= $10\times 12\times 4= 480\ inches^{3}$

Next, if container is placed in its smallest face then it is standing on $(10\times 6)\ inches$ and lets assume height to be "x"

Forming an equation now to express the volume conservation law.

⇒ $480\ inches^{3} = 10\times 6\times x$

⇒$480= 60x$

dividing both side by 60

⇒$x= \frac{480}{60}$

∴ x= 8 inches

Hence, 8 inches from the bottom will the water level reach if the container is placed on its smallest face