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A-small-solid-spherical-ball-of-high-density-is-dropped-in-a-viscous-liquid-Its-journey-in-the-liquid-is-best-described-in-the-following-figure-by-the-curve-

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Question Number 20973 by Tinkutara last updated on 09/Sep/17
A small solid spherical ball of high density is dropped in a viscous liquid. Its journey in the liquid is best described in the following figure by the curve
$$\mathrm{A}\:\mathrm{small}\:\mathrm{solid}\:\mathrm{spherical}\:\mathrm{ball}\:\mathrm{of}\:\mathrm{high} \\ $$$$\mathrm{density}\:\mathrm{is}\:\mathrm{dropped}\:\mathrm{in}\:\mathrm{a}\:\mathrm{viscous}\:\mathrm{liquid}. \\ $$$$\mathrm{Its}\:\mathrm{journey}\:\mathrm{in}\:\mathrm{the}\:\mathrm{liquid}\:\mathrm{is}\:\mathrm{best}\:\mathrm{described} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{figure}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve} \\ $$
Commented by Tinkutara last updated on 09/Sep/17
Answered by dioph last updated on 09/Sep/17
F = m(d^2 x/dt^2 ) Let b ≡ 6πηR mg − b(dx/dt) = m(d^2 x/dt^2 ) (d^2 x/dt^2 ) + (b/m) (dx/dt) = g x(t) = K_1 t + K_2 + K_3 e^(−bt/m) K_1 = ((mg)/b) (dx/dt)∣_0 = 0 ⇔ ((mg)/b) − (b/m)K_3 = 0 ⇒ K_3 = ((m^2 g)/b^2 ) x(0) = 0 ⇔ K_2 + K_3 = 0 ⇒ K_2 = ((−m^2 g)/b^2 ) x(t) = ((mg)/b)(t−(m/b)(1−e^(−bt/m) )) (dx/dt) = ((mg)/b)(1−e^(−bt/m) ) In steady−state, v = ((mg)/b) (constant) So answer should be curve C
$${F}\:=\:{m}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} } \\ $$$$\mathrm{Let}\:{b}\:\equiv\:\mathrm{6}\pi\eta{R} \\ $$$${mg}\:−\:{b}\frac{{dx}}{{dt}}\:=\:{m}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} } \\ $$$$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\:+\:\frac{{b}}{{m}}\:\frac{{dx}}{{dt}}\:=\:{g} \\ $$$${x}\left({t}\right)\:=\:{K}_{\mathrm{1}} {t}\:+\:{K}_{\mathrm{2}} \:+\:{K}_{\mathrm{3}} {e}^{−{bt}/{m}} \\ $$$${K}_{\mathrm{1}} \:=\:\frac{{mg}}{{b}} \\ $$$$\frac{{dx}}{{dt}}\mid_{\mathrm{0}} \:=\:\mathrm{0}\:\Leftrightarrow\:\frac{{mg}}{{b}}\:−\:\frac{{b}}{{m}}{K}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:{K}_{\mathrm{3}} \:=\:\frac{{m}^{\mathrm{2}} {g}}{{b}^{\mathrm{2}} } \\ $$$${x}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:\Leftrightarrow\:{K}_{\mathrm{2}} \:+\:{K}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:{K}_{\mathrm{2}} \:=\:\frac{−{m}^{\mathrm{2}} {g}}{{b}^{\mathrm{2}} } \\ $$$${x}\left({t}\right)\:=\:\frac{{mg}}{{b}}\left({t}−\frac{{m}}{{b}}\left(\mathrm{1}−{e}^{−{bt}/{m}} \right)\right) \\ $$$$\frac{{dx}}{{dt}}\:=\:\frac{{mg}}{{b}}\left(\mathrm{1}−{e}^{−{bt}/{m}} \right) \\ $$$$\mathrm{In}\:\mathrm{steady}−\mathrm{state},\:{v}\:=\:\frac{{mg}}{{b}}\:\left(\mathrm{constant}\right) \\ $$$$\mathrm{So}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{curve}\:\mathrm{C} \\ $$
Commented by Tinkutara last updated on 09/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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