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 \(v^{\prime}\) will be radial and hence \(v^{\prime}\) will be function of \(r^{\prime}\) (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 \(r^{\prime}.\)

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:

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

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

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

\[\begin{aligned} \left\{F^{\prime}(t)\right\}_{r^{\prime}=R} &=\frac{d}{d t}\left(R^{2} V\right)=\frac{d}{d t}\left(R^{2} \frac{d \mathrm{R}}{d t}\right)=\frac{d}{d t}\left(\frac{R}{2} \cdot \frac{d R^{2}}{d t}\right) \\ &=\frac{R}{2} \frac{d^{2} R^{2}}{d t^{2}}+\frac{1}{2} \frac{d R^{2}}{d t} \frac{d R}{d t}=\frac{R}{2} \frac{d^{2} R^{2}}{d t^{2}}+R\left(\frac{d R}{d t}\right)^{2} \end{aligned}\]

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

\[\begin{aligned} P&=\Pi+\frac{1}{2} \rho\left[\frac{2}{R}\left\{\frac{R}{2} \frac{d^{2} R^{2}}{d t^{2}}+R\left(\frac{d R}{d t}\right)^{2}\right\}-\left(\frac{d R}{d t}\right)^{2}\right] \\ &=\Pi+\frac{1}{2} \rho\left[\frac{d^{2} R^{2}}{d t^{2}}+\left(\frac{d R}{d t}\right)^{2}\right] \end{aligned}\]

Hence, proved.

Let \(A B\) and \(A^{\prime} B^{\prime}\) be the ends of the conical pipe such that \(A^{\prime} B^{\prime}=d\) and \(A B=D .\) Let \(\rho_{1}\) and \(\rho_{2}\) be densities of the stream at \(A^{\prime} B^{\prime}\) and \(A B\). By principle of conservation of mass, the mass of the stream that enters the end \(A B\) and leaves at the end \(A^{\prime} B^{\prime}\) 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 \(\frac{v}{V}=\frac{D^{2}}{d^{2}} \times \frac{\rho_{2}}{\rho_{1}} \ldots (1)\)

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

Given that \(\quad p / \rho=k \quad\) so that

\[d p=k d \rho \ldots (3)\]

\(\therefore\) (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

\(\begin{array}{lllll}\text {When } & q=v, & \rho=\rho_{1} & \text { and } & \text { when } & q=V, & \rho=\rho_{2} .\end{array}\) Hence (4) yields
\(k \log \rho_{1}+v^{2} / 2=C \quad\) and \(\quad k \log \rho_{2}+V^{2 /} 2=C\)
Subtracting, \(k\left(\log \rho_{2}-\log \rho_{1}\right)+\left(V^{2}-v^{2}\right) / 2=0\)

\(\implies \log \left(\rho_{2} / \rho_{1}\right)=\left(v^{2}-V^{2}\right) 2 k \quad\) or \(\quad \rho_{2} / \rho_{1}=e^{\left(v^{2}-V^{2}\right) 2 k}\)
Using (5), (1) reduces to \(v / V=\left(D^{2} / d^{2}\right) \times e^{\left(v^{2}-V^{2}\right) / 2 k}\)

Hence, proved.

\(Method I:-\) At any time \(t,\) let \(v^{\prime}\) be the velocity at distance \(r^{\prime}\) from the centre. Again, let \(I\) be the radius of the cavity and \(v\) its velocity. Then the equation of continuity yields \(r^{\prime 2} v^{\prime}=r^{2} v\) When the radius of the cavity is \(r\), then \(\begin{aligned} \text { Kinetic energy } &=\int_{r}^{\infty} \frac{1}{2}\left(4 \pi r^{\prime 2} \rho d r^{\prime}\right) \cdot v^{\prime 2} \quad\left[\because \text { Kinetic energy }=\frac{1}{2} \times \text { mass } \times(\text { velocity })^{2}\right] \\ &=2 \pi \rho r^{4} v^{2} \int_{r}^{\infty} \frac{d r^{\prime}}{r^{\prime 2}}, \text { using }(1) \\ &=2 \pi \rho r^{3} v^{2} \end{aligned}\) The initial kinetic energy is zero. Let \(V\) be the work function (or force potential) due to external forces. Then, we have \(-\frac{\partial V}{\partial r^{\prime}}=\frac{\mu}{r^{\prime 3 / 2}} \quad \text { so that } \quad V=\frac{2 \mu}{r^{1 / 2}}\) \(\therefore\) the work done \(=\int_{r}^{c} V d m, d m\) being the elementary mass \(=\int_{r}^{c}\left(\frac{2 \mu}{r^{\prime 1 / 2}}\right) \cdot 4 \pi r^{\prime 2} d r^{\prime} \rho=8 \pi \mu \rho \int_{T}^{c} r^{\prime 3 / 2} d r^{\prime}=\frac{16}{5} \pi \rho \mu\left(c^{5 / 2}-r^{5 / 2}\right)\) We now use energy equation, namely, Increase in kinetic energy \(=\) work done This \(\Rightarrow \quad 2 \pi \rho r^{3} v^{2}-0=(16 / 5) \times \pi \rho \mu\left(c^{5 / 2}-r^{52}\right)\) \(\therefore\) \(v=\frac{d r}{d t}=-\left(\frac{8 \mu}{5}\right)^{1 / 2} \frac{\left(c^{5 / 2}-r^{5 / 2}\right)^{1 / 2}}{r^{3 / 2}}\) wherein negative sign is taken because \(r\) decreases as \(t\) increases. Let \(T\) be the time of filling up the cavity. Then (2) gives \(\int_{0}^{T} d t=-\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{C}^{0} \frac{r^{3 / 2} d r}{\sqrt{\left(c^{5 / 2}-r^{5 / 2}\right)}} \quad \text { or } \quad T=\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{0}^{c} \frac{r^{3 / 2} d r}{\sqrt{\left(c^{5 / 2}-r^{5 / 2}\right)}}\) \(\begin{array}{l} \text { Put } \quad r^{5 / 2}=c^{5 / 2} \sin ^{2} \theta \quad \text { so that } \quad(5 / 2) \times r^{3 / 2} d r=2 c^{5 / 2} \sin \theta \cos \theta d \theta \\ \therefore \quad T=\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{0}^{\pi / 2} \frac{4}{5} c^{5 / 4} \sin \theta d \theta=\left(\frac{2}{5 \mu}\right)^{1 / 2} c^{5 / 4} \end{array}\) 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 \(v^{\prime}\) will be radial and hence \(v^{\prime}\) will be function of \(r^{\prime}\) (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 \(\begin{aligned} r^{\prime 2} v^{\prime}=F(t) &=r^{2} v \\ & \frac{\partial v^{\prime}}{\partial t}=\frac{F^{\prime}(t)}{r^{\prime 2}} . \end{aligned}\) From (1), The equation of motion is of \(\begin{array}{c} \frac{\partial v^{\prime}}{\partial t}+v^{\prime} \frac{\partial v^{\prime}}{\partial r^{\prime}}=-\frac{\mu}{r^{3 / 2}}-\frac{1}{\rho} \frac{\partial p}{\partial r^{\prime}} \\ \frac{F^{\prime}(t)}{r^{\prime 2}}+\frac{\partial}{\partial r^{\prime}}\left(\frac{1}{2} v^{\prime 2}\right)=-\frac{\mu}{r^{\prime 3 / 2}}-\frac{1}{\rho} \frac{\partial p}{\partial r^{\prime}}, \text { using }(2) \end{array}\) Integrating (3) with respect to \(r^{\prime}\), we have \(-\frac{F^{\prime}(t)}{r^{\prime}}+\frac{1}{2} v^{\prime 2}=\frac{2 \mu}{r^{1 / 2}}-\frac{p}{\rho}+C, C\) being an arbitrary constant When \(r^{\prime}=\infty, v^{\prime}=0, p=0 .\) So from \((4), C=0 .\) Then (4) becomes \(-\frac{F^{\prime}(t)}{r^{\prime}}+\frac{1}{2} v^{\prime 2}=\frac{2 \mu}{r^{\prime 1 / 2}}-\frac{p}{\rho}\) Now when \(r^{\prime}=r, v^{\prime}=v\) and \(p=0 .\) So (5) reduces to \(-\frac{F^{\prime}(t)}{r}+\frac{1}{2} v^{\prime 2}=\frac{2 \mu}{r^{1 / 2}}\) Now, (1) \(\Rightarrow \quad F(t)=r^{2} v \quad \Rightarrow \quad F^{\prime}(t)=2 r v(d r / d t)+r^{2}(d v / d t)\) or \(F^{\prime}(t)=2 r v \frac{d r}{d t}+r^{2} \frac{d v}{d r} \frac{d r}{d t}=2 r v^{2}+r^{2} v \frac{d v}{d r}, \quad\) as \(\quad \frac{d r}{d t}=v\) Hence (6) gives

\(-\frac{1}{r}\left[2 r v^{2}+r^{2} v \frac{d v}{d r}\right]+\frac{v^{2}}{2}=\frac{2 \mu}{r^{1 / 2}} \quad \text { or } \quad r v \frac{d v}{d r}+\frac{3}{2} v^{2}=-\frac{2 \mu}{r^{1 / 2}}\) Multiplying both sides by \(2 r^{2},\) the above equation can be written as \(2 r^{3} v d v+3 r^{2} v^{2} d r=-4 \mu r^{3 / 2} d r \quad \text { or } \quad d\left(r^{3} v^{2}\right)=-4 \mu r^{3 / 2} d r\) 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 \(\begin{array}{c} r^{3} v^{2}=(8 \mu / 5) \times\left(c^{5 / 2}-r^{5 / 2}\right) \\ v=\frac{d r}{d t}=-\left(\frac{8 \mu}{5}\right)^{1 / 2}\left(\frac{c^{5 / 2}-r^{5 / 2}}{r^{3}}\right)^{1 / 2} \end{array}\) taking negative sign for \(d r / d t\) since velocity increases as \(r\) decreases. Let \(T\) be the time of filling up the cavity, then \(T=-\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{c}^{0} \frac{r^{3 / 2} d r}{\left(c^{5 / 2}-r^{5 / 2}\right)^{1 / 2}}\) Let \(r^{5 / 2}=c^{5 / 2} \sin ^{2} \theta\) so that \((5 / 2) \times r^{3 / 2} d r=c^{5 / 2} \sin \theta \cos \theta d \theta\) or \(\begin{array}{c} T=\frac{4}{5}\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{0}^{\pi / 2} \frac{c^{5 / 2} \sin \theta \cos \theta}{c^{5 / 4} \cos \theta} d \theta=\frac{4 c^{5 / 4}}{5}\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{0}^{\pi / 2} \sin \theta d \theta \\ T=(2 / 5 \mu)^{1 / 2} \times c^{5 / 4} \end{array}\)

\(Method I:-\) At any time \(t,\) let \(v^{\prime}\) be the velocity at distance \(r^{\prime}\) from the centre. Again, let \(I\) be the radius of the cavity and \(v\) its velocity. Then the equation of continuity yields \(r^{\prime 2} v^{\prime}=r^{2} v\) When the radius of the cavity is \(r\), then \(\begin{aligned} \text { Kinetic energy } &=\int_{r}^{\infty} \frac{1}{2}\left(4 \pi r^{\prime 2} \rho d r^{\prime}\right) \cdot v^{\prime 2} \quad\left[\because \text { Kinetic energy }=\frac{1}{2} \times \text { mass } \times(\text { velocity })^{2}\right] \\ &=2 \pi \rho r^{4} v^{2} \int_{r}^{\infty} \frac{d r^{\prime}}{r^{\prime 2}}, \text { using }(1) \\ &=2 \pi \rho r^{3} v^{2} \end{aligned}\) The initial kinetic energy is zero. Let \(V\) be the work function (or force potential) due to external forces. Then, we have \(-\frac{\partial V}{\partial r^{\prime}}=\frac{\mu}{r^{\prime 3 / 2}} \quad \text { so that } \quad V=\frac{2 \mu}{r^{1 / 2}}\) \(\therefore\) the work done \(=\int_{r}^{c} V d m, d m\) being the elementary mass \(=\int_{r}^{c}\left(\frac{2 \mu}{r^{\prime 1 / 2}}\right) \cdot 4 \pi r^{\prime 2} d r^{\prime} \rho=8 \pi \mu \rho \int_{T}^{c} r^{\prime 3 / 2} d r^{\prime}=\frac{16}{5} \pi \rho \mu\left(c^{5 / 2}-r^{5 / 2}\right)\) We now use energy equation, namely, Increase in kinetic energy \(=\) work done This \(\Rightarrow \quad 2 \pi \rho r^{3} v^{2}-0=(16 / 5) \times \pi \rho \mu\left(c^{5 / 2}-r^{52}\right)\) \(\therefore\) \(v=\frac{d r}{d t}=-\left(\frac{8 \mu}{5}\right)^{1 / 2} \frac{\left(c^{5 / 2}-r^{5 / 2}\right)^{1 / 2}}{r^{3 / 2}}\) wherein negative sign is taken because \(r\) decreases as \(t\) increases. Let \(T\) be the time of filling up the cavity. Then (2) gives \(\int_{0}^{T} d t=-\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{C}^{0} \frac{r^{3 / 2} d r}{\sqrt{\left(c^{5 / 2}-r^{5 / 2}\right)}} \quad \text { or } \quad T=\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{0}^{c} \frac{r^{3 / 2} d r}{\sqrt{\left(c^{5 / 2}-r^{5 / 2}\right)}}\) \(\begin{array}{l} \text { Put } \quad r^{5 / 2}=c^{5 / 2} \sin ^{2} \theta \quad \text { so that } \quad(5 / 2) \times r^{3 / 2} d r=2 c^{5 / 2} \sin \theta \cos \theta d \theta \\ \therefore \quad T=\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{0}^{\pi / 2} \frac{4}{5} c^{5 / 4} \sin \theta d \theta=\left(\frac{2}{5 \mu}\right)^{1 / 2} c^{5 / 4} \end{array}\) 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 \(v^{\prime}\) will be radial and hence \(v^{\prime}\) will be function of \(r^{\prime}\) (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 \(\begin{aligned} r^{\prime 2} v^{\prime}=F(t) &=r^{2} v \\ & \frac{\partial v^{\prime}}{\partial t}=\frac{F^{\prime}(t)}{r^{\prime 2}} . \end{aligned}\) From (1), The equation of motion is of \(\begin{array}{c} \frac{\partial v^{\prime}}{\partial t}+v^{\prime} \frac{\partial v^{\prime}}{\partial r^{\prime}}=-\frac{\mu}{r^{3 / 2}}-\frac{1}{\rho} \frac{\partial p}{\partial r^{\prime}} \\ \frac{F^{\prime}(t)}{r^{\prime 2}}+\frac{\partial}{\partial r^{\prime}}\left(\frac{1}{2} v^{\prime 2}\right)=-\frac{\mu}{r^{\prime 3 / 2}}-\frac{1}{\rho} \frac{\partial p}{\partial r^{\prime}}, \text { using }(2) \end{array}\) Integrating (3) with respect to \(r^{\prime}\), we have \(-\frac{F^{\prime}(t)}{r^{\prime}}+\frac{1}{2} v^{\prime 2}=\frac{2 \mu}{r^{1 / 2}}-\frac{p}{\rho}+C, C\) being an arbitrary constant When \(r^{\prime}=\infty, v^{\prime}=0, p=0 .\) So from \((4), C=0 .\) Then (4) becomes \(-\frac{F^{\prime}(t)}{r^{\prime}}+\frac{1}{2} v^{\prime 2}=\frac{2 \mu}{r^{\prime 1 / 2}}-\frac{p}{\rho}\) Now when \(r^{\prime}=r, v^{\prime}=v\) and \(p=0 .\) So (5) reduces to \(-\frac{F^{\prime}(t)}{r}+\frac{1}{2} v^{\prime 2}=\frac{2 \mu}{r^{1 / 2}}\) Now, (1) \(\Rightarrow \quad F(t)=r^{2} v \quad \Rightarrow \quad F^{\prime}(t)=2 r v(d r / d t)+r^{2}(d v / d t)\) or \(F^{\prime}(t)=2 r v \frac{d r}{d t}+r^{2} \frac{d v}{d r} \frac{d r}{d t}=2 r v^{2}+r^{2} v \frac{d v}{d r}, \quad\) as \(\quad \frac{d r}{d t}=v\) Hence (6) gives

\(-\frac{1}{r}\left[2 r v^{2}+r^{2} v \frac{d v}{d r}\right]+\frac{v^{2}}{2}=\frac{2 \mu}{r^{1 / 2}} \quad \text { or } \quad r v \frac{d v}{d r}+\frac{3}{2} v^{2}=-\frac{2 \mu}{r^{1 / 2}}\) Multiplying both sides by \(2 r^{2},\) the above equation can be written as \(2 r^{3} v d v+3 r^{2} v^{2} d r=-4 \mu r^{3 / 2} d r \quad \text { or } \quad d\left(r^{3} v^{2}\right)=-4 \mu r^{3 / 2} d r\) 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 \(\begin{array}{c} r^{3} v^{2}=(8 \mu / 5) \times\left(c^{5 / 2}-r^{5 / 2}\right) \\ v=\frac{d r}{d t}=-\left(\frac{8 \mu}{5}\right)^{1 / 2}\left(\frac{c^{5 / 2}-r^{5 / 2}}{r^{3}}\right)^{1 / 2} \end{array}\) taking negative sign for \(d r / d t\) since velocity increases as \(r\) decreases. Let \(T\) be the time of filling up the cavity, then \(T=-\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{c}^{0} \frac{r^{3 / 2} d r}{\left(c^{5 / 2}-r^{5 / 2}\right)^{1 / 2}}\) Let \(r^{5 / 2}=c^{5 / 2} \sin ^{2} \theta\) so that \((5 / 2) \times r^{3 / 2} d r=c^{5 / 2} \sin \theta \cos \theta d \theta\) or \(\begin{array}{c} T=\frac{4}{5}\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{0}^{\pi / 2} \frac{c^{5 / 2} \sin \theta \cos \theta}{c^{5 / 4} \cos \theta} d \theta=\frac{4 c^{5 / 4}}{5}\left(\frac{5}{8 \mu}\right)^{1 / 2} \int_{0}^{\pi / 2} \sin \theta d \theta \\ T=(2 / 5 \mu)^{1 / 2} \times c^{5 / 4} \end{array}\)