A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and...
Question 1 Report
A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times thr radius of its base.
(a) Illustrate this information in a diagram.
(b) If the volume of the reservoir is \(333\frac{1}{3}\pi m^{3}\), calculate, correct to the nearest whole number, the :
(I) volume of the hemisphere ; (II) Total surface area of the reservoir. [Take \(\pi = \frac{22}{7}\)].
Answer Details
(a) Diagram: A cone sitting on top of a hemisphere, sharing the same circular base of radius \(r\). The flat face of the hemisphere coincides with the base of the cone, and the cone height is \(h = 6r\).
(b) Let the common radius be \(r\), so \(h = 6r\).
\(V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi r^2(6r) = 2\pi r^3\)
\(V_{\text{hemisphere}} = \tfrac{2}{3}\pi r^3\)
Total: \(2\pi r^3 + \tfrac{2}{3}\pi r^3 = \tfrac{8}{3}\pi r^3 = 333\tfrac{1}{3}\pi = \tfrac{1000}{3}\pi\).
\(8r^3 = 1000 \ \Rightarrow\ r^3 = 125 \ \Rightarrow\ r = 5\text{ m}\) (so \(h = 30\text{ m}\)).
(I) Volume of hemisphere: \(\tfrac{2}{3}\pi r^3 = \tfrac{2}{3}\times\tfrac{22}{7}\times 125 = \dfrac{5500}{21} = 261.9 \approx \mathbf{262\text{ m}^3}\).
(II) Total surface area \(=\) curved surface of cone \(+\) curved surface of hemisphere (the joining faces are internal).
Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{25 + 900} = \sqrt{925} = 30.41\text{ m}\).
\(\pi r l = \tfrac{22}{7}\times 5\times 30.41 = 477.9\text{ m}^2\); \(2\pi r^2 = 2\times\tfrac{22}{7}\times 25 = 157.1\text{ m}^2\).
Total \(= 477.9 + 157.1 = 635.0 \approx \mathbf{635\text{ m}^2}\).
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