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Question Number 152101 by otchereabdullai@gmail.com last updated on 25/Aug/21

$$\mathrm{Mr}\mathrm{Bonsu}\mathrm{an}\mathrm{engineer}\mathrm{walked}\mathrm{round}$$$$\mathrm{a}\mathrm{cylindrical}\mathrm{container}4\mathrm{m}\mathrm{high}\mathrm{once}$$$$\mathrm{keeping}\mathrm{a}\mathrm{constant}\mathrm{distance}\mathrm{of}1\mathrm{m}\mathrm{from}$$$$\mathrm{the}\mathrm{container}.\mathrm{If}\mathrm{he}\mathrm{walked}\mathrm{with}\mathrm{a}$$$$\mathrm{speed}\mathrm{of}{3\mathrm{kmh}}^{-1}\mathrm{for}\mathrm{three}\mathrm{minutes},$$$$\mathrm{calculate}\mathrm{to}\mathrm{the}\mathrm{nearest}\mathrm{whole}\mathrm{number}$$$$\mathrm{the}:$$$$\left(\mathrm{i}\right)\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$\left(\mathrm{ii}\right)\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{container}$$

Answered by Olaf_Thorendsen last updated on 25/Aug/21

$$2\pi {(r+1)}^2=3\mathrm{km}.{\mathrm{h}}^{-1}\times 3\min$$$$2\pi {(r+1)}^2=\frac{3000}{60}\times 3=150$$$$r=\sqrt{\frac{75}{\pi }}-1\approx 4m$$$$\mathrm{V}=\pi r^{2}h=\pi {(\sqrt{\frac{75}{\pi }}-1)}^2\times 4\approx 190{\mathrm{m}}^3$$$$$$

Commented by otchereabdullai@gmail.com last updated on 25/Aug/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}!$$

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