Answer:
1.5 m
Step-by-step explanation:
Given that:
The diameter of the cylinder = 1.4 m
Then the radius will be = 1.4 m/ 2 = 0.7 m = 7dm
Similarly, the height of the container = 3.6 m = 36 dm
Suppose the height of the inverted cone = h_c and the height of the cylinder tube = h_d
Then the tank height = h_c + h_d
36 m = h_c + h_d ------ (1)
However, the volume of the water in the cone can be computed as:
[tex]= \dfrac{1}{3} \times \pi \times r ^ 2 \times h_c[/tex]
Similarly, the volume of the water in the cylinder tube is:
[tex]= \pi \times r ^ 2 \times h_d[/tex]
The volume of water in the container = 2.464 liters
Thus;
The volume of water in the cone + volume of water in the tube = volume of water
[tex]( \dfrac{1}{3} \times \pi \times r ^ 2 \times h_c )+( \pi \times r ^ 2 \times h_d) = 2464[/tex]
[tex]\pi \times r ^ 2 ( \dfrac{1}{3} \times h_c + h_d) = 2464[/tex]
[tex]\pi \times 7^ 2 ( \dfrac{1}{3} \times h_c + 30 - h_c) = 2464[/tex]
[tex]154 ( -\dfrac{2}{3} h_c + 30 ) = 2464[/tex]
[tex]( -\dfrac{2}{3} h_c + 30 ) = \dfrac{2464}{154}[/tex]
[tex]( -\dfrac{2}{3} h_c + 30 ) =16[/tex]
[tex]( -\dfrac{2}{3} h_c ) = 16-30[/tex]
[tex]-\dfrac{2}{3} h_c =-14[/tex]
[tex]h_c =-14 \div- \dfrac{2}{3}[/tex]
[tex]h_c =-7 \times-3[/tex]
[tex]h_c =21[/tex]
From equation (1):
36 m = h_c + h_d
h_d = 36 - h_c
h_d = 36 - 21
h_d = 15 dm
Therefore, the height of the cylinder tube is 15 dm = 1.5 m
