Answer:
W=96040π N*m
Explanation:
Weight water is [tex]=9800 N/m^3[/tex]
Weight layer is [tex]9800*v N/m^3[/tex]
The v is given by
[tex]r=2m[/tex]
[tex]v=\pi*r^2*y' =(4\pi*y')m^3[/tex]
so weight layer is
[tex]9800*4\pi*y'=39200\pi[/tex]
The distance layer moved is
[tex]d=7-y[/tex]
So the work is
[tex]W=\int\limits^7_0{(7-y)*39200\pi } \, dy[/tex]
[tex]W=39200*[7*y-\frac{y^2}{2}] |0,7[/tex]
[tex]W=39200*[49-24.5]=39200\pi *24.5[/tex]
[tex]W=960400\pi N*m[/tex]
The required work done is [tex]940800\pi[/tex]
Given,
Radius=2m
Height=6m
As the tank is below the ground level by 1m
The center of the mass of the water is,
[tex]H=\frac{6}{2}+1\\=4 m[/tex]
The formula of the volume of the cyllindrical tank is,
[tex]V=\pi r^2h[/tex]
Now, substitute the given values into the above formula we get,
[tex]V=\pi r^2h\\=\pi(2)^2\times6\\=24\pi m^3[/tex]
The given density is 9800[tex]N/m^3[/tex]
[tex]weight=density \times volume\\=9800\times24\pi\\=235200\pi N[/tex]
So, the required work done is,
[tex]Work=mgh\\=235200\pi \times4\\=940800 \pi[/tex]
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