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Question Number 29687 by ajfour last updated on 11/Feb/18

Commented by ajfour last updated on 11/Feb/18

$$Findthefinalvelocitygainedbya$$$$tankofmassM,initiallycompletely$$$$filledwithwaterandatrest,asit$$$$slipsonafrictionlessgroundowing$$$$toasmallopeningatitsbaseof$$$$radiusa,throughwhichitdrains$$$$out.\left(Assumedensityofwater\mathit{\rho }\right).$$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{isnt}\mathrm{area}\mathrm{of}\mathrm{hole}\mathrm{given}?$$

Commented by ajfour last updated on 11/Feb/18

$$yes\pi a^2\left(pleasereadquestionagain\right).$$

Commented by ajfour last updated on 11/Feb/18

$$whatsthefinalspeedoftank$$$$bythetimey\to 0.$$

Answered by mrW2 last updated on 12/Feb/18

$$M_W=massofwateratbegin=\pi R^{2}h\rho$$$$atwaterdepthy:$$$$totalmass:m=M+M_W\frac{y}{h}$$$$\Rightarrow \frac{dm}{dt}=\frac{M_W}{h}\times \frac{dy}{dt}$$$$\left(relative\right)speedofoutgoingwater:$$$$v=\sqrt{2gy}$$$$$$$${let}^{\prime }ssayspeedoftank=u$$$$-v\frac{dm}{dt}=m\frac{du}{dt}$$$$-v\times \frac{M_W}{h}\times \frac{dy}{dt}=m\frac{du}{dy}\times \frac{dy}{dt}$$$$-\frac{M_W}{h}v=m\frac{du}{dy}$$$$-\frac{M_W}{h}\sqrt{2gy}=(M+\frac{M_W}{h}y)\frac{du}{dy}$$$$-\frac{M_W}{h}\times \frac{\sqrt{2gy}}{M+\frac{M_W}{h}y}dy=du$$$$-\sqrt{2g}\times \frac{\sqrt{y}}{(\frac{hM}{M_W}+y)}dy=du$$$$withk=\frac{hM}{M_W}$$$$-\sqrt{2g}\times {\int }_h^0\frac{\sqrt{y}}{(k+y)}dy={\int }_0^{u_{max}}du$$$$\sqrt{2g}\times 2\sqrt{k}{[\sqrt{\frac{y}{k}}-{\tan }^{-1}\sqrt{\frac{y}{k}}]}_0^h=u_{max}$$$$u_{max}=2\sqrt{2gk}(\sqrt{\frac{h}{k}}-{\tan }^{-1}\sqrt{\frac{h}{k}})$$$$u_{max}=2\sqrt{\frac{2ghM}{M_W}}(\sqrt{\frac{M_W}{M}}-{\tan }^{-1}\sqrt{\frac{M_W}{M}})$$$$with\lambda =\frac{M_W}{M}$$$$\Rightarrow u_{max}=2\sqrt{\frac{2gh}{\lambda }}(\sqrt{\lambda }-{\tan }^{-1}\sqrt{\lambda })$$$$$$$$Example:\lambda =\frac{M_W}{M}=5,h=1m$$$$\Rightarrow u_{max}=2\sqrt{\frac{20}{5}}(\sqrt{5}-{\tan }^{-1}\sqrt{5})=4.3m/s$$$$====================$$$$u\left(y\right)=2\sqrt{2gk}(\sqrt{\frac{h}{k}}-\sqrt{\frac{y}{k}}-{\tan }^{-1}\sqrt{\frac{h}{k}}+{\tan }^{-1}\sqrt{\frac{y}{k}})$$$$\pi a^{2}vdt=-\pi R^{2}dy$$$$\Rightarrow \frac{dy}{dt}=-\frac{a^2}{R^2}v=-\frac{a^2\sqrt{2gy}}{R^2}$$$$\Rightarrow \frac{dy}{\sqrt{y}}=-\frac{a^2\sqrt{2g}}{R^2}dt$$$$\Rightarrow {\int }_h^y\frac{dy}{\sqrt{y}}=-\frac{a^2\sqrt{2g}}{R^2}{\int }_0^{t}dt$$$$\Rightarrow 2(\sqrt{y}-\sqrt{h})=-\frac{a^2\sqrt{2g}}{R^2}\times t$$$$\Rightarrow \sqrt{y}=\sqrt{h}-\frac{a^2\sqrt{2g}}{2R^2}\times t=\sqrt{h}-pt$$$$\Rightarrow y=h-2\sqrt{h}pt+p^2{t}^2$$$$y=0\Rightarrow t=\frac{\sqrt{h}}{p}=\frac{R^2}{a^2}\sqrt{\frac{2h}{g}}=t_1\left(at{t}_{1}tankisempty\right)$$$$$$$$\Rightarrow u\left(t\right)=2\sqrt{2gk}[\frac{pt}{\sqrt{k}}-{\tan }^{-1}\sqrt{\frac{h}{k}}+{\tan }^{-1}(\sqrt{\frac{h}{k}}-\frac{pt}{\sqrt{k}})]$$$$with\eta =\frac{p}{\sqrt{k}}=\frac{a^2}{R^2}\sqrt{\frac{2g\lambda }{h}}$$$$\Rightarrow u\left(t\right)=2\sqrt{\frac{2gh}{\lambda }}[\eta t-{\tan }^{-1}\sqrt{\lambda }+{\tan }^{-1}(\sqrt{\lambda }-\eta t)]$$

Commented by mrW2 last updated on 12/Feb/18

Is any solution for this question given?

Commented by ajfour last updated on 12/Feb/18

$$\mathbb{SUPERB}!Sir,Energymethod$$$$leadstoacomplicateddifferential$$$$equation,iguess.$$$$Yourmethodis\mathcal{E}xcellent,Sir.$$

Commented by ajfour last updated on 12/Feb/18

$$Nosir,notfrombook.$$

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