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  • The figure shows a hollow cube of side 'a' of volume V. There is a small chamber of volume <img alt="\frac{V}{4}" src="https://learn.careers360.com/latex-image/?%5Cfrac%7BV%7D%7B4%7D"

The figure shows a hollow cube of side 'a' of volume V. There is a small chamber of volume \frac{V}{4}  in the cube as shown. This chamber is completely filled by m kg of water.  Water leaks through a hole H. Then the work done by gravity in this process is W=\frac{x}{8}mga (assuming that the complete water finally lies at the bottom of the cube is). Then what is the value of x

:

Option: 1

4


Option: 2

3


Option: 3

5


Option: 4

1


Answers (1)

best_answer

The center of mass of water when kept in chamber lies at a height  \frac{a}{2}+\frac{a}{4}=\frac{3 a}{4}$  from the bottom of the cube.
Hence the initial potential energy of water =m g \frac{3 a}{4}$

The same amount of water then falls inside the cube. Let the height upto which water fills the cube be h.
since the volume of water does not change,
a \times \frac{a}{2} \times \frac{a}{2}=a \times a \times h$ $\Longrightarrow h=\frac{a}{4}$
Hence the center of mass of water in the cube =\frac{1}{2} \times \frac{a}{4}=\frac{a}{8}$

The gravitational potential of water finally is m g \frac{a}{8}$
Hence the work done by gravity on water = Loss in potential energy,
=m g a\left(\frac{3}{4}-\frac{1}{8}\right)$ $=\frac{5}{8} m g a$.

so x=5

Posted by

manish

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