Volume of cone, frustum and hemisphere

Question Sample Titled 'Volume of cone, frustum and hemisphere'

Figure (a) shows a solid right circular cone of base radius ${144}$ $\text{cm}$ and ${73}$ $\text{cm}$ .

${73}\text{cm}$${144}\text{cm}$Figure (a)Figure (b)

 (a) Find the volume of the circular cone in terms of $\pi$ . (2 marks) (b) A hemispherical vessel of radius ${150}$ $\text{cm}$ is held vertically on a horizontal surface. The vessel is fully filled with alcohol. (5 marks) (i) Find the volume of the alcohol in the vessel in terms of $\pi$ . (i) The circular cone is now held vertically in the vessel as shown in Figure (b). Jamie claims that the volume of the alcohol remaining in the vessel is greater than ${5.6}$ $\text{m}^{{3}}$ . Do you agree? Explain your answer.

 (a) The required volume $=\dfrac{{1}}{{3}}\pi{\left({144}\right)}^{{2}}{\left({73}\right)}$ 1M  $={504576}\pi$ $\text{cm}^{{3}}$ 1A  (b) (i) The required volume $=\dfrac{{4}}{{3}}\pi{\left({150}\right)}^{{3}}\cdot\dfrac{{1}}{{2}}$ 1M  $={2250000}\pi$ $\text{cm}^{{3}}$ 1A  (b) (ii) The height of the frustum under the surface of the alcohol  $=\sqrt{{{150}^{{2}}-{144}^{{2}}}}$ 1M  $={42}$ $\text{cm}$ The volume of the alcohol remaining in the vessel  $={2250000}\pi-{\left({504576}\pi{\left({1}-{\left(\dfrac{{{73}-{42}}}{{73}}\right)}^{{3}}\right)}\right)}$ 1M  $=\dfrac{{255121228}}{{143}}\pi$ $\text{cm}^{{3}}$  $={5.604804025476225}$ $\text{m}^{{3}}$  $>{5.6}$ $\text{m}^{{3}}$ Thus, the claim is agreed. 1A

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