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Question Number 47471 by ajfour last updated on 10/Nov/18

Commented by ajfour last updated on 10/Nov/18

$$Findy\left(\theta \right).$$

Commented by peter frank last updated on 25/Dec/21

$$\mathrm{what}\mathrm{is}\mathrm{name}\mathrm{of}\mathrm{this}\mathrm{app}$$

Answered by MrW3 last updated on 11/Nov/18

$$volumeofwaterV=\pi R^{2}h$$$${\theta }_m={\tan }^{-1}\frac{h}{R}$$$$$$$$case1:\theta ⩽{\theta }_m$$$$bottomofcantainerissubmerged.$$$$\pi R^2(h_1+h_2)/2=\pi R^{2}h$$$$\Rightarrow h_1+h_2=2h$$$$h_2-h_1=2R\tan \theta$$$$\Rightarrow h_2=h+R\tan \theta$$$$y=h_2\cos \theta =(h+R\tan \theta )\cos \theta$$$$\Rightarrow y\left(\theta \right)=R\sin \theta +h\cos \theta with\theta ⩽{\theta }_m$$$$$$$$case2:\theta >{\theta }_m$$$$bottomofcontainerisnotcompletely$$$$submerged.$$$$h_3=R(1-\cos \alpha )$$$$h_2=h_3\tan \theta =R(1-\cos \alpha )\tan \theta$$$$A=\frac{R^2}{2}(2\alpha -\sin 2\alpha )=R^2(\alpha -\sin \alpha \cos \alpha )$$$$V=\frac{{Ah}_2}{2}=\frac{R^3(\alpha -\sin \alpha \cos \alpha )(1-\cos \alpha )\tan \theta }{2}=\pi R^{2}h$$$$\Rightarrow \frac{R(\alpha -\sin \alpha \cos \alpha )(1-\cos \alpha )\tan \theta }{2}=\pi h$$$$\Rightarrow (\alpha -\sin \alpha \cos \alpha )(1-\cos \alpha )=\frac{2\pi h}{R\tan \theta }$$$$\Rightarrow \alpha =\alpha \left(\theta \right)\left(onlynumerically\right)$$$$y=h_2\cos \theta$$$$\Rightarrow y\left(\theta \right)=R[1-\cos \alpha \left(\theta \right)]\sin \theta$$

Commented by MrW3 last updated on 10/Nov/18

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