Answer:
Base radius is increasing by 2.88 m after 5 minutes
Step-by-step explanation:
Given:-
Height of cone (h) =2[tex]\times[/tex] base diameter(d)
[tex]h=2\times d[/tex]
Machine starts dumping sand at the rate of 20 [tex]m^{3}/min[/tex]
So,
Volume of cone in 1 min = 20 [tex]m^{3}[/tex]
Now, Volume of cone in 5 mins = Volume of cone in 1 min [tex]\times 5[/tex]
Volume of cone in 5 mins = 20 [tex]\times 5[/tex]
Volume of cone in 5 mins = 100 [tex]m^{3}[/tex] ---------(equation 1)
Now, Let base diameter of cone = b
height of the cone = h
formula for volume of cone is,
Volume of cone (V) = [tex]\pi\times r^{2} \times\frac{h}{3}[/tex]
[tex]V=\pi\times( \frac{d}{2} )^{2}\times(\frac{2\times d}{3})[/tex] ----(since r=[tex]\frac{d}{2}[/tex])
[tex]V=\pi\times\frac{d^{2} }{4} \times \frac{2d}{3}[/tex]
[tex]V=\frac{\pi\times d^{3} }{6}[/tex] ---------(equation 2)
Now substituting equation 1 in equation 2,
[tex]100=\frac{\pi\times d^{3} }{6}[/tex]
[tex]d^{3}=\frac{100\times 6}{3.14}[/tex] ---------(as [tex]\pi=3.14[/tex])
[tex]d^{3} =\frac{600}{3.14}[/tex]
[tex]d^{3} =191.08[/tex]
By cube rooting both the sides we get,
[tex]\sqrt[3]{d} =\sqrt[3]{191.08}[/tex]
[tex]d=5.76 \ m[/tex] ----------------(diameter)
[tex]as\ r=\frac{d}{2}[/tex]
[tex]r=\frac{5.76}{2}[/tex]
[tex]r=2.88\ m[/tex] ---------------(radius of base at 5 mins)
Therefore base radius is increasing by 2.88 m after 5 minutes
