A tank in the shape of an inverted right circular cone has height 4 meters and radius 2 meters. It is filled with 3 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ=1020 kg/m3.δ=1020 kg/m3. Your answer must include the correct units.
Question
Answer:
The required work to empty the the tank is 247401 J.
Explanation:
Given that,
A tank is in the shape of right circular cone.
The radius of the tank= 2 m
The height of the tank = 4 m
The relation between the radius and the height is
[tex]\frac{r}{h}=\frac{2}{4}[/tex]
[tex]\Rightarrow r= \frac12 h[/tex]
Let h be the height of hot of chocolate any time t.
The volume of a cone is [tex]= \frac 13 \pi r^2 h[/tex]
[tex]=\frac13 \pi (\frac h2)^2h[/tex]
[tex]=\frac16 \pi h^3[/tex]
The volume of the chocolate is [tex]=\frac16 \pi h^3[/tex]
The mass of the chocolate is(M)= Density × Volume
[tex]=1020(\frac16 \pi h^3)[/tex] Kg
[tex]=170\pi h^3[/tex] kg
[tex]\frac{dM}{dh}= 170\pi (3h^2)[/tex]
[tex]\Rightarrow dM=510\pi h^2\ dh[/tex]
Work done = Force × displacement
= Mass × acceleration×displacement
Here acceleration= acceleration due to gravity = 9.8 m/s²
The displacement when the hot chocolate level is h is = (4-h)
Work done (dw) [tex]=(510\pi h^2 )(9.8)(4-h)dh[/tex] [tex]=4998\pi h^2 (4-h)dh[/tex]
Work done = W [tex]=\int_0^34998\pi h^2 (4-h)dh[/tex]
[tex]=\int_0^34998\pi [4h^2-h^3]dh[/tex]
[tex]=4998\pi [4\frac{h^3}{3}-\frac{h^4}{4}]_0^3[/tex]
[tex]=4998\pi [(4\frac{3^3}{3}-\frac{3^4}{4})-(4\frac{0^3}{3}-\frac{0^4}{4})][/tex]
=247401 J
The required work to empty the the tank is 247401 J.