PYQs

Here the motion of the fluid will take place in such a manner so that each element of the fluid moves towards the centre. Hence the free surface would be spherical. Thus the fluid velocity $v^{\mathrm{\prime }}$ will be radial and hence $v^{\mathrm{\prime }}$ will be function of $r^{\mathrm{\prime }}$ (the radial distance from the centre of the sphere which is taken as origin), and time $t$ only. Let $p$ be pressure at a distance $r^{\mathrm{\prime }}.$

Let $P$ be the pressure on the surface of the sphere of radius $R$ and $V$ be the velocity there. Then the equation of continuity is:

$r^{\mathrm{\prime }2}{v}^{\mathrm{\prime }}=R^{2}V=F(t)\dots (1)$ From (1) $\frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }t}=\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }2}}\dots (2)$ Again equation of motion is: $\frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }t}+v^{\mathrm{\prime }}\frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }{r}^{\mathrm{\prime }}}=-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }{r}^{\mathrm{\prime }}}$ $⟹$ $-\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }2}}+\frac{\mathrm{\partial }}{\mathrm{\partial }{r}^{\mathrm{\prime }}}\left(\frac{1}{2}{v}^{\mathrm{\prime }2}\right)=-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }{r}^{\mathrm{\prime }}},$ (using (2)) … (3)

Integrating with respect to $r^{\mathrm{\prime }}$, (3) reduces to: $-\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }}}+\frac{1}{2}{v}^{\mathrm{\prime }2}=-\frac{p}{\rho }+C$, $C$ being an arbitrary constant When $r^{\mathrm{\prime }}=\mathrm{\infty },$ then $v^{\mathrm{\prime }}=0$ and $p=\mathrm{\Pi }$ so that $C=\mathrm{\Pi }/\rho .$ Then, we get $-\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }}}+\frac{1}{2}{v}^{\mathrm{\prime }2}=\frac{\mathrm{\Pi }-p}{\rho }$ $⟹p=\mathrm{\Pi }+\frac{1}{2}\rho [2\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }}}-v^{\mathrm{\prime }2}]\dots (4)$ But $p=P$ and $v^{\mathrm{\prime }}=V$ when $r^{\mathrm{\prime }}=\mathrm{R}$. Hence (4) gives:

$\mathrm{P}=\mathrm{\Pi }+\frac{1}{2}\rho [\frac{2}{\mathrm{R}}{\{F^{\mathrm{\prime }}(t)\}}_{r^{\mathrm{\prime }}=R}-{\mathrm{V}}^2]\dots (5)$ Also $V=dR/dt.$ Hence using $(1),$ we have

Using the above values of $V$ and ${\{F^{\mathrm{\prime }}(t)\}}_{r^{\mathrm{\prime }}=R},(5)$ reduces to

Hence, proved.

Let $AB$ and $A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}$ be the ends of the conical pipe such that $A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}=d$ and $AB=D.$ Let ${\rho }_1$ and ${\rho }_2$ be densities of the stream at $A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}$ and $AB$. By principle of conservation of mass, the mass of the stream that enters the end $AB$ and leaves at the end $A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}$ must be the same. Hence the equation of continuity is $\pi (d/2{)}^{2}v{\rho }_1=\pi (D/2{)}^{2}V{\rho }_2$

so that $\frac{v}{V}=\frac{D^2}{d^2}\times \frac{{\rho }_2}{{\rho }_1}\dots (1)$

By Bernoulli’s theorem (in absence of external forces like gravity), we have
$\int \frac{dp}{\rho }+\frac{1}{2}{q}^2=C\dots (2)$

Given that $p/\rho =k$ so that

$\therefore$ (2) reduces to
$k\int \frac{d\rho }{\rho }+\frac{1}{2}{q}^2=C$ (using (3))
Integrating above, we get
$k\log \rho +q^2/2=C$, $C$ being an arbitrary constant

$\begin{array}{lllllll}\text{When }& q=v,& \rho ={\rho }_1& \text{ and }& \text{ when }& q=V,& \rho ={\rho }_2.\end{array}$ Hence (4) yields
$k\log {\rho }_1+v^2/2=C$ and $k\log {\rho }_2+V^{2/}2=C$
Subtracting, $k(\log {\rho }_2-\log {\rho }_1)+(V^2-v^2)/2=0$

$⟹\log \left({\rho }_2/{\rho }_1\right)=(v^2-V^2)2k$ or ${\rho }_2/{\rho }_1=e^{(v^2-V^2)2k}$
Using (5), (1) reduces to $v/V=\left(D^2/d^2\right)\times e^{(v^2-V^2)/2k}$

Hence, proved.

$MethodI:-$ At any time $t,$ let $v^{\mathrm{\prime }}$ be the velocity at distance $r^{\mathrm{\prime }}$ from the centre. Again, let $I$ be the radius of the cavity and $v$ its velocity. Then the equation of continuity yields $r^{\mathrm{\prime }2}{v}^{\mathrm{\prime }}=r^{2}v$ When the radius of the cavity is $r$, then $\begin{array}{rl}\text{ Kinetic energy }& ={\int }_r^{\mathrm{\infty }}\frac{1}{2}\left(4\pi r^{\mathrm{\prime }2}\rho d{r}^{\mathrm{\prime }}\right)\cdot v^{\mathrm{\prime }2}[\because \text{ Kinetic energy }=\frac{1}{2}\times \text{ mass }\times (\text{ velocity }{)}^2]\\ & =2\pi \rho r^4{v}^2{\int }_r^{\mathrm{\infty }}\frac{d{r}^{\mathrm{\prime }}}{r^{\mathrm{\prime }2}},\text{ using }(1)\\ & =2\pi \rho r^3{v}^2\end{array}$ The initial kinetic energy is zero. Let $V$ be the work function (or force potential) due to external forces. Then, we have $-\frac{\mathrm{\partial }V}{\mathrm{\partial }{r}^{\mathrm{\prime }}}=\frac{\mu }{r^{\mathrm{\prime }3/2}}\text{ so that }V=\frac{2\mu }{r^{1/2}}$ $\therefore$ the work done $={\int }_r^{c}Vdm,dm$ being the elementary mass $={\int }_r^c\left(\frac{2\mu }{r^{\mathrm{\prime }1/2}}\right)\cdot 4\pi r^{\mathrm{\prime }2}d{r}^{\mathrm{\prime }}\rho =8\pi \mu \rho {\int }_T^c{r}^{\mathrm{\prime }3/2}d{r}^{\mathrm{\prime }}=\frac{16}{5}\pi \rho \mu (c^{5/2}-r^{5/2})$ We now use energy equation, namely, Increase in kinetic energy $=$ work done This $\Rightarrow 2\pi \rho r^3{v}^2-0=(16/5)\times \pi \rho \mu (c^{5/2}-r^{52})$ $\therefore$ $v=\frac{dr}{dt}=-{\left(\frac{8\mu }{5}\right)}^{1/2}\frac{{(c^{5/2}-r^{5/2})}^{1/2}}{r^{3/2}}$ wherein negative sign is taken because $r$ decreases as $t$ increases. Let $T$ be the time of filling up the cavity. Then (2) gives ${\int }_0^{T}dt=-{\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_C^0\frac{r^{3/2}dr}{\sqrt{(c^{5/2}-r^{5/2})}}\text{ or }T={\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_0^c\frac{r^{3/2}dr}{\sqrt{(c^{5/2}-r^{5/2})}}$ $\begin{array}{l}\text{ Put }{r}^{5/2}=c^{5/2}{\sin }^2\theta \text{ so that }(5/2)\times r^{3/2}dr=2c^{5/2}\sin \theta \cos \theta d\theta \\ \therefore T={\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_0^{\pi /2}\frac{4}{5}{c}^{5/4}\sin \theta d\theta ={\left(\frac{2}{5\mu }\right)}^{1/2}{c}^{5/4}\end{array}$ Second Method:- Here the motion of the fluid will take place in such a manner so that each element of the fluid moves towards the centre. Hence the free surface would be spherical. Thus the fluid velocity $v^{\mathrm{\prime }}$ will be radial and hence $v^{\mathrm{\prime }}$ will be function of $r^{\mathrm{\prime }}$ (the radial distance from the centre of the sphere which is taken as origin) and time $t$. Also, let $v$ be the velocity at a distance $r$. Then the equation of continuity is $\begin{array}{rl}{r}^{\mathrm{\prime }2}{v}^{\mathrm{\prime }}=F(t)& =r^{2}v\\ & \frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }t}=\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }2}}.\end{array}$ From (1), The equation of motion is of $\begin{array}{c}\frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }t}+v^{\mathrm{\prime }}\frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }{r}^{\mathrm{\prime }}}=-\frac{\mu }{r^{3/2}}-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }{r}^{\mathrm{\prime }}}\\ \frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }2}}+\frac{\mathrm{\partial }}{\mathrm{\partial }{r}^{\mathrm{\prime }}}\left(\frac{1}{2}{v}^{\mathrm{\prime }2}\right)=-\frac{\mu }{r^{\mathrm{\prime }3/2}}-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }{r}^{\mathrm{\prime }}},\text{ using }(2)\end{array}$ Integrating (3) with respect to $r^{\mathrm{\prime }}$, we have $-\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }}}+\frac{1}{2}{v}^{\mathrm{\prime }2}=\frac{2\mu }{r^{1/2}}-\frac{p}{\rho }+C,C$ being an arbitrary constant When $r^{\mathrm{\prime }}=\mathrm{\infty },v^{\mathrm{\prime }}=0,p=0.$ So from $(4),C=0.$ Then (4) becomes $-\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }}}+\frac{1}{2}{v}^{\mathrm{\prime }2}=\frac{2\mu }{r^{\mathrm{\prime }1/2}}-\frac{p}{\rho }$ Now when $r^{\mathrm{\prime }}=r,v^{\mathrm{\prime }}=v$ and $p=0.$ So (5) reduces to $-\frac{F^{\mathrm{\prime }}(t)}{r}+\frac{1}{2}{v}^{\mathrm{\prime }2}=\frac{2\mu }{r^{1/2}}$ Now, (1) $\Rightarrow F(t)=r^{2}v\Rightarrow F^{\mathrm{\prime }}(t)=2rv(dr/dt)+r^2(dv/dt)$ or $F^{\mathrm{\prime }}(t)=2rv\frac{dr}{dt}+r^2\frac{dv}{dr}\frac{dr}{dt}=2r{v}^2+r^{2}v\frac{dv}{dr},$ as $\frac{dr}{dt}=v$ Hence (6) gives

$-\frac{1}{r}[2r{v}^2+r^{2}v\frac{dv}{dr}]+\frac{v^2}{2}=\frac{2\mu }{r^{1/2}}\text{ or }rv\frac{dv}{dr}+\frac{3}{2}{v}^2=-\frac{2\mu }{r^{1/2}}$ Multiplying both sides by $2r^2,$ the above equation can be written as $2r^{3}vdv+3r^2{v}^{2}dr=-4\mu r^{3/2}dr\text{ or }d\left(r^3{v}^2\right)=-4\mu r^{3/2}dr$ Integrating. $r^3{v}^2=-(8\mu /5)r^{5/2}+D,D$ being an arbitrary constant When $r=c,v=0.$ So (7) gives $D=(8\mu /5)c^{5/2}.$ Hence (7) reduces to or $\begin{array}{c}{r}^3{v}^2=(8\mu /5)\times (c^{5/2}-r^{5/2})\\ v=\frac{dr}{dt}=-{\left(\frac{8\mu }{5}\right)}^{1/2}{\left(\frac{c^{5/2}-r^{5/2}}{r^3}\right)}^{1/2}\end{array}$ taking negative sign for $dr/dt$ since velocity increases as $r$ decreases. Let $T$ be the time of filling up the cavity, then $T=-{\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_c^0\frac{r^{3/2}dr}{{(c^{5/2}-r^{5/2})}^{1/2}}$ Let $r^{5/2}=c^{5/2}{\sin }^2\theta$ so that $(5/2)\times r^{3/2}dr=c^{5/2}\sin \theta \cos \theta d\theta$ or $\begin{array}{c}T=\frac{4}{5}{\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_0^{\pi /2}\frac{c^{5/2}\sin \theta \cos \theta }{c^{5/4}\cos \theta }d\theta =\frac{4c^{5/4}}{5}{\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_0^{\pi /2}\sin \theta d\theta \\ T=(2/5\mu {)}^{1/2}\times c^{5/4}\end{array}$

$MethodI:-$ At any time $t,$ let $v^{\mathrm{\prime }}$ be the velocity at distance $r^{\mathrm{\prime }}$ from the centre. Again, let $I$ be the radius of the cavity and $v$ its velocity. Then the equation of continuity yields $r^{\mathrm{\prime }2}{v}^{\mathrm{\prime }}=r^{2}v$ When the radius of the cavity is $r$, then $\begin{array}{rl}\text{ Kinetic energy }& ={\int }_r^{\mathrm{\infty }}\frac{1}{2}\left(4\pi r^{\mathrm{\prime }2}\rho d{r}^{\mathrm{\prime }}\right)\cdot v^{\mathrm{\prime }2}[\because \text{ Kinetic energy }=\frac{1}{2}\times \text{ mass }\times (\text{ velocity }{)}^2]\\ & =2\pi \rho r^4{v}^2{\int }_r^{\mathrm{\infty }}\frac{d{r}^{\mathrm{\prime }}}{r^{\mathrm{\prime }2}},\text{ using }(1)\\ & =2\pi \rho r^3{v}^2\end{array}$ The initial kinetic energy is zero. Let $V$ be the work function (or force potential) due to external forces. Then, we have $-\frac{\mathrm{\partial }V}{\mathrm{\partial }{r}^{\mathrm{\prime }}}=\frac{\mu }{r^{\mathrm{\prime }3/2}}\text{ so that }V=\frac{2\mu }{r^{1/2}}$ $\therefore$ the work done $={\int }_r^{c}Vdm,dm$ being the elementary mass $={\int }_r^c\left(\frac{2\mu }{r^{\mathrm{\prime }1/2}}\right)\cdot 4\pi r^{\mathrm{\prime }2}d{r}^{\mathrm{\prime }}\rho =8\pi \mu \rho {\int }_T^c{r}^{\mathrm{\prime }3/2}d{r}^{\mathrm{\prime }}=\frac{16}{5}\pi \rho \mu (c^{5/2}-r^{5/2})$ We now use energy equation, namely, Increase in kinetic energy $=$ work done This $\Rightarrow 2\pi \rho r^3{v}^2-0=(16/5)\times \pi \rho \mu (c^{5/2}-r^{52})$ $\therefore$ $v=\frac{dr}{dt}=-{\left(\frac{8\mu }{5}\right)}^{1/2}\frac{{(c^{5/2}-r^{5/2})}^{1/2}}{r^{3/2}}$ wherein negative sign is taken because $r$ decreases as $t$ increases. Let $T$ be the time of filling up the cavity. Then (2) gives ${\int }_0^{T}dt=-{\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_C^0\frac{r^{3/2}dr}{\sqrt{(c^{5/2}-r^{5/2})}}\text{ or }T={\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_0^c\frac{r^{3/2}dr}{\sqrt{(c^{5/2}-r^{5/2})}}$ $\begin{array}{l}\text{ Put }{r}^{5/2}=c^{5/2}{\sin }^2\theta \text{ so that }(5/2)\times r^{3/2}dr=2c^{5/2}\sin \theta \cos \theta d\theta \\ \therefore T={\left(\frac{5}{8\mu }\right)}^{1/2}{\int }_0^{\pi /2}\frac{4}{5}{c}^{5/4}\sin \theta d\theta ={\left(\frac{2}{5\mu }\right)}^{1/2}{c}^{5/4}\end{array}$ Second Method:- Here the motion of the fluid will take place in such a manner so that each element of the fluid moves towards the centre. Hence the free surface would be spherical. Thus the fluid velocity $v^{\mathrm{\prime }}$ will be radial and hence $v^{\mathrm{\prime }}$ will be function of $r^{\mathrm{\prime }}$ (the radial distance from the centre of the sphere which is taken as origin) and time $t$. Also, let $v$ be the velocity at a distance $r$. Then the equation of continuity is $\begin{array}{rl}{r}^{\mathrm{\prime }2}{v}^{\mathrm{\prime }}=F(t)& =r^{2}v\\ & \frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }t}=\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }2}}.\end{array}$ From (1), The equation of motion is of $\begin{array}{c}\frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }t}+v^{\mathrm{\prime }}\frac{\mathrm{\partial }{v}^{\mathrm{\prime }}}{\mathrm{\partial }{r}^{\mathrm{\prime }}}=-\frac{\mu }{r^{3/2}}-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }{r}^{\mathrm{\prime }}}\\ \frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }2}}+\frac{\mathrm{\partial }}{\mathrm{\partial }{r}^{\mathrm{\prime }}}\left(\frac{1}{2}{v}^{\mathrm{\prime }2}\right)=-\frac{\mu }{r^{\mathrm{\prime }3/2}}-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }{r}^{\mathrm{\prime }}},\text{ using }(2)\end{array}$ Integrating (3) with respect to $r^{\mathrm{\prime }}$, we have $-\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }}}+\frac{1}{2}{v}^{\mathrm{\prime }2}=\frac{2\mu }{r^{1/2}}-\frac{p}{\rho }+C,C$ being an arbitrary constant When $r^{\mathrm{\prime }}=\mathrm{\infty },v^{\mathrm{\prime }}=0,p=0.$ So from $(4),C=0.$ Then (4) becomes $-\frac{F^{\mathrm{\prime }}(t)}{r^{\mathrm{\prime }}}+\frac{1}{2}{v}^{\mathrm{\prime }2}=\frac{2\mu }{r^{\mathrm{\prime }1/2}}-\frac{p}{\rho }$ Now when $r^{\mathrm{\prime }}=r,v^{\mathrm{\prime }}=v$ and $p=0.$ So (5) reduces to $-\frac{F^{\mathrm{\prime }}(t)}{r}+\frac{1}{2}{v}^{\mathrm{\prime }2}=\frac{2\mu }{r^{1/2}}$ Now, (1) $\Rightarrow F(t)=r^{2}v\Rightarrow F^{\mathrm{\prime }}(t)=2rv(dr/dt)+r^2(dv/dt)$ or $F^{\mathrm{\prime }}(t)=2rv\frac{dr}{dt}+r^2\frac{dv}{dr}\frac{dr}{dt}=2r{v}^2+r^{2}v\frac{dv}{dr},$ as $\frac{dr}{dt}=v$ Hence (6) gives