Answer:
Step-by-step explanation:
Given
Radius of cone [tex]R=3\ m[/tex]
height of cone [tex]H=6\ m[/tex]
Chocolate is filled up to a height of [tex]h=2\ m[/tex]
Work done to lift a body to a height h when F force is applied is given by
[tex]W=Fh[/tex]
An infinitesimally slice of chocolate [tex]\Delta y[/tex] at a height of y has a radius
[tex]r=\dfrac{R}{H}\cdot h[/tex]
Volume of slice is [tex]dV=\pi r^2y[/tex]
[tex]mass =\rho \times dV[/tex]
This mass need to be raised to a height of 6-y which requires a work of
[tex]dW=\rho (\pi r^2ydy)g(6-y)[/tex]
To empty the tank
[tex]W=\int_{0}^{3}\rho \cdot \pi \cdot g\cdot r^2\cdot (6-y)dy[/tex]
at [tex]h=y[/tex]
[tex]r=\dfrac{R}{H}\cdot y[/tex]
[tex]r=\dfrac{3}{6}y[/tex]
[tex]r=\dfrac{y}{2}[/tex]
[tex]W=\int_{0}^{3}\rho \cdot \pi \cdot g\cdot \dfrac{y^2}{4}\cdot (6-y)dy[/tex]
[tex]W=10^3\times \pi \times \dfrac{1}{4}\times (2y^3-\dfrac{1}{4}y^4)_0^3[/tex]
[tex]W=259.80\ kJ[/tex]
