Question-227276

Question Number 227276 by mr W last updated on 11/Jan/26

Commented by mr W last updated on 12/Jan/26

$${A}\:{hemispherical}\:{funnel}\:{is}\:{placed} \\ $$$${tightly}\:{against}\:{the}\:{top}\:{of}\:{a}\:{table}.\: \\ $$$${Water}\:{is}\:{poured}\:{into}\:{it}\:{slowly}\: \\ $$$${through}\:{a}\:{small}\:{hole}\:{located}\:{at}\:{the}\: \\ $$$${highest}\:{point}\:{of}\:{the}\:{funnel}.\:{When}\: \\ $$$${the}\:{water}\:{level}\:{inside}\:{the}\:{funnel}\: \\ $$$${just}\:{reaches}\:{the}\:{hole}\:{the}\:{funnel}\: \\ $$$${begins}\:{to}\:{float}\:{upward}\:{and}\:{the}\: \\ $$$${water}\:{starts}\:{to}\:{flow}\:{out}\:{from}\:{the}\: \\ $$$${bottom}.\:{If}\:{the}\:{radius}\:{of}\:\:{the}\:{funnel}\: \\ $$$${is}\:{R}=\mathrm{10}\:{cm}\:{and}\:{the}\:{density}\:{of}\:{water} \\ $$$${is}\:\rho=\mathrm{1}\:{g}/{cm}^{\mathrm{3}} ,\:{find}\:{the}\:{mass}\:{of}\:{the}\: \\ $$$${funnel}. \\ $$

Answered by Kassista last updated on 13/Jan/26

Critical moment: Buoyancy = Weight of the funnel ∴ ρV_(hemisphere ) g = Mg ⇒ρ((2πR^3 )/3) = M { ((ρ =1g.cm^(−3) )),((R=10 cm)) :}⇒ M = ((2000π)/3) g = ((2π)/3) kg ≈ 2.09kg

$$ \\ $$$${Critical}\:{moment}:\:{Buoyancy}\:=\:{Weight}\:{of}\:{the}\:{funnel} \\ $$$$\therefore\:\rho{V}_{{hemisphere}\:} {g}\:=\:{Mg}\:\Rightarrow\rho\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}\:=\:{M} \\ $$$$\begin{cases}{\rho\:=\mathrm{1}{g}.{cm}^{−\mathrm{3}} }\\{{R}=\mathrm{10}\:{cm}}\end{cases}\Rightarrow\:{M}\:=\:\frac{\mathrm{2000}\pi}{\mathrm{3}}\:{g}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:{kg}\:\approx\:\mathrm{2}.\mathrm{09}{kg} \\ $$$$ \\ $$

Commented by mr W last updated on 13/Jan/26

$${why}\:{is}\:{buoyancy}\:=\rho{g}\:{V}_{{hemisphere}} \:? \\ $$

Answered by fantastic2 last updated on 17/Jan/26

Commented by mr W last updated on 17/Jan/26

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Commented by fantastic2 last updated on 17/Jan/26

h=R(1−sin α) P=hρg=R(1−sin α)ρg Area of disc =2πRcos α×Rdα =2πR^2 cos αdα dF=P×A =2πR^3 ρg(1−sin α)cos αdα dF_⊥ =2πR^3 ρg(1−sin α)cos αsin αdα F=∫_0 ^(π/2) 2πR^3 ρg(1−sin α)cos αsin αdα =2πR^3 ρg(∫_0 ^(π/2) sin αcos αdα−∫_0 ^(π/2) sin^2 αcos αdα) let y=sin α ∴(dy/dα)=cos α ⇒dy=cos αdα sin 0=0 sin (π/2)=1 ∴2πR^3 ρg(∫_0 ^1 y dy−∫_0 ^1 y^2 dy) =2πR^3 ρg((1/2)−(1/3)) =2πR^3 ρg×(1/6) =(1/3)πR^3 ρg Mg=(1/3)πR^3 ρg ∴M=(1/3)πR^3 ρ=(1/3)πR^3

$${h}={R}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right) \\ $$$${P}={h}\rho{g}={R}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)\rho{g} \\ $$$${Area}\:{of}\:{disc} \\ $$$$=\mathrm{2}\pi{R}\mathrm{cos}\:\alpha×{Rd}\alpha \\ $$$$=\mathrm{2}\pi{R}^{\mathrm{2}} \mathrm{cos}\:\alpha{d}\alpha \\ $$$${dF}={P}×{A} \\ $$$$=\mathrm{2}\pi{R}^{\mathrm{3}} \rho{g}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)\mathrm{cos}\:\alpha{d}\alpha \\ $$$${dF}_{\bot} =\mathrm{2}\pi{R}^{\mathrm{3}} \rho{g}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)\mathrm{cos}\:\alpha\mathrm{sin}\:\alpha{d}\alpha \\ $$$${F}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}\pi{R}^{\mathrm{3}} \rho{g}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)\mathrm{cos}\:\alpha\mathrm{sin}\:\alpha{d}\alpha \\ $$$$=\mathrm{2}\pi{R}^{\mathrm{3}} \rho{g}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:\alpha\mathrm{cos}\:\alpha{d}\alpha−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:^{\mathrm{2}} \alpha\mathrm{cos}\:\alpha{d}\alpha\right) \\ $$$${let}\:{y}=\mathrm{sin}\:\alpha \\ $$$$\therefore\frac{{dy}}{{d}\alpha}=\mathrm{cos}\:\alpha \\ $$$$\Rightarrow{dy}=\mathrm{cos}\:\alpha{d}\alpha \\ $$$$\mathrm{sin}\:\mathrm{0}=\mathrm{0} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{2}}=\mathrm{1} \\ $$$$\therefore\mathrm{2}\pi{R}^{\mathrm{3}} \rho{g}\left(\int_{\mathrm{0}} ^{\mathrm{1}} {y}\:{dy}−\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}} \:{dy}\right) \\ $$$$=\mathrm{2}\pi{R}^{\mathrm{3}} \rho{g}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}\pi{R}^{\mathrm{3}} \rho{g}×\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \rho{g} \\ $$$${Mg}=\frac{\mathrm{1}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \rho{g} \\ $$$$\therefore{M}=\frac{\mathrm{1}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \rho=\frac{\mathrm{1}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \\ $$$$ \\ $$

Answered by mr W last updated on 17/Jan/26

Commented by mr W last updated on 17/Jan/26

weight of funnel = buoyancy forec = water weight displaced = ρg×hitched volume =ρg(πR^2 ×R−((2πR^3 )/3)) =((πR^3 ρg)/3) ⇒mass of funnel =((πR^3 ρ)/3)

$${weight}\:{of}\:{funnel}\: \\ $$$$=\:{buoyancy}\:{forec} \\ $$$$=\:{water}\:{weight}\:{displaced}\: \\ $$$$=\:\rho{g}×{hitched}\:{volume} \\ $$$$=\rho{g}\left(\pi{R}^{\mathrm{2}} ×{R}−\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}\right) \\ $$$$=\frac{\pi{R}^{\mathrm{3}} \rho{g}}{\mathrm{3}} \\ $$$$\Rightarrow{mass}\:{of}\:{funnel}\:=\frac{\pi{R}^{\mathrm{3}} \rho}{\mathrm{3}} \\ $$

Answered by mr W last updated on 17/Jan/26

Commented by mr W last updated on 17/Jan/26

at the moment as the funnel starts to float upward, the contact force between funnel and ground is zero. p=ρgR total pressure = total weight of water and funnel πR^2 ×ρgR=((2πR^3 ρg)/3)+M_(funnel) g ⇒M_(funnel) =((πR^3 ρ)/3)

$${at}\:{the}\:{moment}\:{as}\:{the}\:{funnel}\:{starts} \\ $$$${to}\:{float}\:{upward},\:{the}\:{contact}\:{force} \\ $$$${between}\:{funnel}\:{and}\:{ground}\:{is}\:{zero}. \\ $$$${p}=\rho{gR} \\ $$$${total}\:{pressure}\: \\ $$$$=\:{total}\:{weight}\:{of}\:{water}\:{and}\:{funnel} \\ $$$$\pi{R}^{\mathrm{2}} ×\rho{gR}=\frac{\mathrm{2}\pi{R}^{\mathrm{3}} \rho{g}}{\mathrm{3}}+{M}_{{funnel}} {g} \\ $$$$\Rightarrow{M}_{{funnel}} =\frac{\pi{R}^{\mathrm{3}} \rho}{\mathrm{3}} \\ $$

Commented by fantastic2 last updated on 17/Jan/26

$${WooooW}! \\ $$

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