Types of Flow, Equation of Continuity and Flow Rate Questions in English

(A) Given: Diameter of first pipe $d_A = 2 \text{ cm}$,so radius $r_A = 1 \text{ cm}$.
Diameter of second pipe $d_B = 4 \text{ cm}$,so radius $r_B = 2 \text{ cm}$.
According to the equation of continuity,the product of the cross-sectional area and the velocity of flow is constant $(A_1v_1 = A_2v_2)$.
Therefore,$\frac{v_A}{v_B} = \frac{A_B}{A_A} = \frac{\pi (r_B)^2}{\pi (r_A)^2}$.
Substituting the values: $\frac{v_A}{v_B} = \left( \frac{2}{1} \right)^2 = 4$.
Thus,$v_A = 4v_B$. The velocity in the $2 \text{ cm}$ pipe is $4$ times the velocity in the $4 \text{ cm}$ pipe.

(C) For an incompressible liquid, the volume flow rate entering the junction must equal the volume flow rate leaving the junction (Equation of Continuity).
$A_{in} v_{in} = A_1 v_1 + A_2 v_2$
From the figure, the inlet area is $A$ with velocity $v_1 = 3 \ m/s$. The outlet areas are $A$ and $1.5A$ with velocities $v_2 = 1.5 \ m/s$ and $v$ respectively.
$A \times 3 = A \times 1.5 + 1.5A \times v$
Dividing both sides by $A$:
$3 = 1.5 + 1.5v$
$1.5 = 1.5v$
$v = 1 \ m/s$.

(D) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area $A$ and the velocity $v$ of the fluid remains constant at all points along the tube,i.e.,$A_1 v_1 = A_2 v_2$.
Since the tube is cylindrical,the cross-sectional area is uniform throughout,meaning $A_1 = A_2$.
Therefore,$v_1 = v_2$ must hold true regardless of the orientation of the tube (horizontal or vertical).
This principle is based on the conservation of mass,which applies to all three cases.

(D) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of flow remains constant at any point along the pipe.
$A_1 V_1 = A_2 V_2$
Given:
$A_1 = 4.20 \; cm^2$
$V_1 = 5.18 \; ms^{-1}$
$A_2 = 7.60 \; cm^2$
Substituting the values into the equation:
$4.20 \times 5.18 = 7.60 \times V_2$
$V_2 = \frac{4.20 \times 5.18}{7.60}$
$V_2 = \frac{21.756}{7.60}$
$V_2 \approx 2.86 \; ms^{-1}$
Therefore,the speed of flow at the lower level is $2.86 \; ms^{-1}$.

(A) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of the fluid remains constant at all points along the pipe.
$A_1 v_1 = A_2 v_2$
Here,$A_1$ and $A_2$ are the cross-sectional areas at the entry and exit,and $v_1$ and $v_2$ are the velocities at the entry and exit,respectively.
Since the cross-section is circular,$A = \pi r^2$,where $r$ is the radius.
Therefore,$\pi r_1^2 v_1 = \pi r_2^2 v_2$.
This simplifies to $\frac{v_1}{v_2} = \frac{r_2^2}{r_1^2} = \left( \frac{r_2}{r_1} \right)^2$.
Given the ratio of radii at entry and exit is $r_1 : r_2 = 3 : 2$,we have $\frac{r_2}{r_1} = \frac{2}{3}$.
Substituting this value: $\frac{v_1}{v_2} = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.
Thus,the ratio of velocities at the entry and exit is $4:9$.

(B) According to the equation of continuity for an incompressible,non-viscous fluid in steady flow,the mass flow rate remains constant throughout the tube.
The volume flow rate (or discharge) is given by the product of the cross-sectional area and the velocity of the fluid.
Therefore,$A_1 v_1 = A_2 v_2$.
To find the ratio of the speeds $v_1/v_2$,we rearrange the equation:
$v_1/v_2 = A_2/A_1$.
Thus,the correct option is $B$.

(C) In a streamline flow,the velocity of a fluid particle at any given point is constant over time.
This means that every particle that passes through a specific point $P$ will have the same velocity (both magnitude and direction) as the previous particle that passed through that same point $P$.
Since kinetic energy $K = \frac{1}{2}mv^2$ depends on the speed $v$ of the particle,and all particles passing through point $P$ have the same velocity,they must also have the same kinetic energy.
Therefore,option $C$ is the correct statement.

(C) According to the equation of continuity,the product of the cross-sectional area and the velocity of an incompressible fluid remains constant throughout the flow: $A_1 v_1 = A_2 v_2$.
Here,$A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
Substituting this into the equation: $\frac{\pi d_1^2}{4} v_1 = \frac{\pi d_2^2}{4} v_2$.
This simplifies to: $d_1^2 v_1 = d_2^2 v_2$.
Given $d_1 = 4 \text{ cm}$,$v_1 = 3 \text{ m/s}$,and $d_2 = 2 \text{ cm}$.
Substituting the values: $(4)^2 \times 3 = (2)^2 \times v_2$.
$16 \times 3 = 4 \times v_2$.
$48 = 4 v_2$.
$v_2 = 12 \text{ m/s}$.

(A) When water falls from a tap,it accelerates due to gravity as it moves downwards.
According to the equation of continuity,$A_{1} v_{1} = A_{2} v_{2}$,where $A$ is the cross-sectional area and $v$ is the velocity of the liquid.
Since the velocity $v$ increases as the water falls,the cross-sectional area $A$ must decrease to keep the flow rate constant.
Therefore,the area of the water stream decreases.

(D) According to the equation of continuity for an incompressible fluid,the volume flow rate remains constant at all points in the pipe.
Therefore,the volume of fluid entering per second at point $A$ must equal the volume of fluid exiting per second at point $B$.
Let $v_A$ be the velocity at point $A$ and $v_B$ be the velocity at point $B$.
The area of cross-section at $A$ is $A_A = \pi(2r)^2 = 4\pi r^2$.
The area of cross-section at $B$ is $A_B = \pi(r)^2 = \pi r^2$.
Using the equation of continuity: $A_A v_A = A_B v_B$.
Substituting the given values: $(4\pi r^2) v = (\pi r^2) v_B$.
Solving for $v_B$: $v_B = \frac{4\pi r^2 v}{\pi r^2} = 4v$.

(A) The mass of the liquid flowing out per unit time is given by $\frac{dm}{dt} = \rho A v$.
The kinetic energy $K$ of the liquid is $K = \frac{1}{2} m v^2$.
The rate at which kinetic energy is imparted is the power $P = \frac{dK}{dt} = \frac{1}{2} \left( \frac{dm}{dt} \right) v^2$.
Substituting the value of $\frac{dm}{dt}$:
$P = \frac{1}{2} (\rho A v) v^2 = \frac{1}{2} A \rho v^3$.

(A) The mass of the liquid passing through the pipe per unit time is given by the rate of mass flow: $\frac{dm}{dt} = \rho A V$,where $\rho = d$ is the density.
Thus,$\frac{dm}{dt} = dAV$.
The kinetic energy $K$ of a mass $m$ moving with speed $V$ is $K = \frac{1}{2} m V^2$.
The kinetic energy imparted per unit time is $\frac{dK}{dt} = \frac{1}{2} V^2 \frac{dm}{dt}$.
Substituting the value of $\frac{dm}{dt}$,we get $\frac{dK}{dt} = \frac{1}{2} V^2 (dAV) = \frac{1}{2} AdV^3$.

(A) According to the equation of continuity,the product of the cross-sectional area $(A)$ and the velocity of the fluid $(v)$ remains constant for an incompressible fluid in steady flow: $A_1 v_1 = A_2 v_2$.
Given radii $r_1 = 2 \, cm$ and $r_2 = 4 \, cm$.
The area is given by $A = \pi r^2$.
Therefore,$\pi (r_1)^2 v_1 = \pi (r_2)^2 v_2$.
The ratio of velocities is $\frac{v_1}{v_2} = \frac{(r_2)^2}{(r_1)^2} = \left( \frac{4}{2} \right)^2 = (2)^2 = 4$.
Thus,the ratio of the flow velocities is $4:1$.

(C) According to the equation of continuity for an incompressible fluid,the total mass flow rate entering the junction must equal the total mass flow rate leaving the junction.
Since the fluid is incompressible,the volume flow rate is conserved:
$A_1 v_1 = A_2 v_2 + A_3 v_3$
From the figure,the inlet area is $A$ with velocity $v_1 = 3 \ m/s$.
The outlet areas are $A$ with velocity $v_2 = 1.5 \ m/s$ and $1.5A$ with velocity $v$.
Substituting these values into the equation:
$A \times 3 = A \times 1.5 + 1.5A \times v$
Dividing both sides by $A$:
$3 = 1.5 + 1.5v$
$1.5 = 1.5v$
$v = 1 \ m/s$

(C) According to the equation of continuity,the volume flow rate of the liquid remains constant throughout the tube.
The volume flow rate inside the cylindrical tube is given by $A_1 v_1 = \pi R^2 v$.
The total volume flow rate through the $n$ fine holes is given by $A_2 v_2 = n (\pi r^2) v_{ejection}$.
Equating the two flow rates:
$\pi R^2 v = n \pi r^2 v_{ejection}$.
Solving for the speed of ejection $(v_{ejection})$:
$v_{ejection} = \frac{\pi R^2 v}{n \pi r^2} = \frac{v R^2}{n r^2}$.

(B) Turbulent flow is characterized by high velocity,specifically when the velocity of the fluid exceeds the critical velocity $(v_c)$. In this state,the flow becomes irregular or chaotic,and fluid particles move across layers,causing mixing. Conversely,when the velocity is less than the critical velocity,the flow is laminar (streamline),which is smooth and orderly. Therefore,having a velocity less than the critical velocity is a characteristic of laminar flow,not turbulent flow. The correct option is $B$.

(C) The correct explanation is based on the equation of continuity. For an incompressible fluid in streamline flow,the volume flow rate remains constant at every cross-section of the flow.
According to the equation of continuity: $A_1 V_1 = A_2 V_2 = \text{constant}$,where $A$ is the cross-sectional area and $V$ is the velocity of the fluid.
As the water falls vertically under gravity,its velocity $V$ increases as it moves downwards $(V = \sqrt{2gh})$.
Since the product $A \times V$ must remain constant,if the velocity $V$ increases,the cross-sectional area $A$ must decrease to maintain the constant flow rate. Therefore,the water column tapers as it falls.

(D) According to the equation of continuity,the volume flow rate of water remains constant.
Let $A_1$ be the cross-sectional area of the pipe and $v_1$ be the velocity of water in the pipe.
Let $A_2$ be the total cross-sectional area of the $100$ holes and $v_2$ be the velocity of water coming out of the holes.
$A_1 v_1 = 100 \times A_{hole} \times v_2$
Given:
Diameter of pipe $D_1 = 2 \ cm = 2 \times 10^{-2} \ m$,so radius $r_1 = 1 \times 10^{-2} \ m$.
Velocity $v_1 = 3 \ ms^{-1}$.
Diameter of each hole $d = 0.05 \ cm = 0.05 \times 10^{-2} \ m$,so radius $r = 0.025 \times 10^{-2} \ m$.
Substituting the values:
$\pi (1 \times 10^{-2})^2 \times 3 = 100 \times \pi (0.025 \times 10^{-2})^2 \times v_2$
$10^{-4} \times 3 = 100 \times (2.5 \times 10^{-4})^2 \times v_2$
$3 \times 10^{-4} = 100 \times 6.25 \times 10^{-8} \times v_2$
$3 \times 10^{-4} = 6.25 \times 10^{-6} \times v_2$
$v_2 = \frac{3 \times 10^{-4}}{6.25 \times 10^{-6}} = \frac{300}{6.25} = 48 \ ms^{-1}$.

(A) Given: Diameter of pipe $P$,$d_{p} = 2 \times 10^{-2} \ m$,so radius $r_{p} = 10^{-2} \ m$.
Diameter of pipe $Q$,$d_{q} = 4 \times 10^{-2} \ m$,so radius $r_{q} = 2 \times 10^{-2} \ m$.
According to the equation of continuity,the product of cross-sectional area and velocity is constant for an incompressible fluid: $A_{p}v_{p} = A_{q}v_{q}$.
Therefore,the velocity is inversely proportional to the cross-sectional area: $\frac{v_{p}}{v_{q}} = \frac{A_{q}}{A_{p}}$.
Since $A = \pi r^{2}$,we have $\frac{v_{p}}{v_{q}} = \frac{\pi r_{q}^{2}}{\pi r_{p}^{2}} = \left(\frac{r_{q}}{r_{p}}\right)^{2}$.
Substituting the values: $\frac{v_{p}}{v_{q}} = \left(\frac{2 \times 10^{-2}}{10^{-2}}\right)^{2} = (2)^{2} = 4$.
Thus,$v_{p} = 4v_{q}$. The velocity of water in pipe $P$ is $4$ times that of $Q$.

(B) Let $A_1 = 1 \ cm^2$ and $A_2 = 0.5 \ cm^2$ be the cross-sectional areas at the tap and at a distance $h = 10 \ cm$ below it,respectively.
From the equation of continuity,$A_1 v_1 = A_2 v_2$,we have $1 \cdot v_1 = 0.5 \cdot v_2$,which implies $v_2 = 2v_1$.
Using Bernoulli's equation between the two points: $P_0 + \frac{1}{2}\rho v_1^2 + \rho gh = P_0 + \frac{1}{2}\rho v_2^2$.
Substituting $v_2 = 2v_1$: $gh + \frac{1}{2}v_1^2 = \frac{1}{2}(2v_1)^2 = 2v_1^2$.
Thus,$gh = \frac{3}{2}v_1^2$,so $v_1 = \sqrt{\frac{2gh}{3}}$.
Given $g = 980 \ cm/s^2$ and $h = 10 \ cm$,$v_1 = \sqrt{\frac{2 \cdot 980 \cdot 10}{3}} = \sqrt{\frac{19600}{3}} \approx 80.83 \ cm/s$.
Volumetric flow rate $Q = A_1 v_1 = 1 \ cm^2 \cdot 80.83 \ cm/s = 80.83 \ cm^3/s$.
Converting to $litre/min$: $Q = 80.83 \times 10^{-3} \ litre/s = 80.83 \times 10^{-3} \times 60 \ litre/min \approx 4.85 \ litre/min$.
Rounding to the nearest option,we get $4.9 \ litre/min$.

(C) Let the velocity at the tap be $v_1 = 0.4 \ m/s$ and the diameter be $D_1 = 8 \times 10^{-3} \ m$.
Let the velocity at a distance $h = 0.2 \ m$ below the tap be $v_2$ and the diameter be $D_2$.
Using the equation of motion $v_2^2 = v_1^2 + 2gh$,we get:
$v_2 = \sqrt{(0.4)^2 + 2 \times 10 \times 0.2} = \sqrt{0.16 + 4} = \sqrt{4.16} \approx 2.04 \ m/s$.
From the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A = \frac{\pi D^2}{4}$.
Thus,$D_1^2 v_1 = D_2^2 v_2$.
$D_2 = D_1 \sqrt{\frac{v_1}{v_2}} = (8 \times 10^{-3}) \sqrt{\frac{0.4}{2.04}} \approx (8 \times 10^{-3}) \sqrt{0.196} \approx (8 \times 10^{-3}) \times 0.443 \approx 3.54 \times 10^{-3} \ m$.
Rounding to the nearest provided option,the value is close to $3.6 \times 10^{-3} \ m$.

(C) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity remains constant,i.e.,$A_{1}v_{1} = A_{2}v_{2}$.
Here,the cross-sectional area $A$ is given by the product of width $W$ and depth $d$. Since the width $W$ is constant,we have:
$A_{1} = W \times 4 \ m$
$A_{2} = W \times 8 \ m$
Given $v_{1} = 12 \ m/s$,we substitute these values into the continuity equation:
$(W \times 4) \times 12 = (W \times 8) \times v_{2}$
Dividing both sides by $W$ (since $W \neq 0$):
$4 \times 12 = 8 \times v_{2}$
$48 = 8 \times v_{2}$
$v_{2} = \frac{48}{8} = 6 \ m/s$.
Thus,the velocity at the $8 \ m$ depth is $6 \ m/s$.

(A) The rate of flow (volume flow rate) is given by the equation $Q = A \times v$,where $A$ is the cross-sectional area and $v$ is the average velocity.
Given:
$A = 0.25\,m^2$
$Q = 100\,cm^3/s = 100 \times 10^{-6}\,m^3/s = 10^{-4}\,m^3/s$
Rearranging the formula for velocity:
$v = \frac{Q}{A} = \frac{10^{-4}}{0.25} = 4 \times 10^{-4}\,m/s$
To convert the velocity from $m/s$ to $mm/s$,multiply by $1000$:
$v = 4 \times 10^{-4} \times 10^3\,mm/s = 0.4\,mm/s$.

(D) According to the principle of continuity, for a streamline flow of an incompressible, non-viscous fluid through a tube of non-uniform cross-section, the volume rate of flow (also known as the discharge rate, $Q$) remains constant at every cross-section of the tube.
The equation of continuity is given by $A_1 v_1 = A_2 v_2 = \text{constant}$, where $A$ is the cross-sectional area and $v$ is the velocity of the fluid.
Since the fluid is incompressible and the flow is steady (streamline), the amount of fluid entering any section of the tube in a given time must equal the amount of fluid leaving that section.
Therefore, the rate of flow of the liquid is the same at both points $M$ and $N$.

(D) According to the equation of continuity for an incompressible fluid in streamline flow,the volume flow rate (or discharge) $Q_v = A \cdot v$ remains constant throughout the tube.
This means that the volume of liquid entering any cross-section per unit time must equal the volume of liquid leaving any other cross-section per unit time.
Therefore,the rate of liquid flow is the same at all points in the tube,including points $P$ and $Q$.

(A) The volume of air passing through the area $A$ in time $t$ is given by the product of the area,velocity,and time: $V = A \times v \times t$.
Since the density of the air is $\rho$,the mass $m$ of the air is given by the product of density and volume: $m = \rho \times V$.
Substituting the expression for volume,we get: $m = \rho \times (A \times v \times t) = Av\rho t$.

(A) When water falls from a tap,it accelerates due to gravity as it moves downward.
According to the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A$ is the cross-sectional area and $v$ is the velocity.
Since the velocity $v$ increases as the water falls,the cross-sectional area $A$ must decrease to keep the flow rate constant.
Therefore,the area of the water stream decreases.

(D) In streamline flow,the velocity of a fluid particle at any given point is constant over time.
Since the velocity $v$ at a specific point is constant,the momentum $p = mv$ of any particle arriving at that point will be the same,assuming the particles have the same mass $m$.
Similarly,the kinetic energy $K = \frac{1}{2}mv^2$ of any particle passing through that specific point will also be the same.
Therefore,both statements $(A)$ and $(C)$ are correct.

(D) Turbulent flow is characterized by chaotic,irregular fluid motion.
In turbulent flow,the fluid particles move in random paths,which significantly enhances the mixing of different layers of the fluid.
This increased mixing leads to a higher rate of transfer of momentum and energy between fluid layers compared to laminar flow.
Additionally,the internal friction caused by these chaotic motions dissipates kinetic energy into heat.
Therefore,the statement that turbulent flow decreases the rate of transfer of momentum and energy is incorrect.

(A) According to the equation of continuity,the total flow rate $Q_{total}$ is equal to the sum of the flow rates in the two branches.
$Q_{total} = A_1 v_p + A_2 v_Q$
Given:
$Q_{total} = 0.1\, m^3/sec$
$A_1 = A/3 = 10^{-2}/3\, m^2$
$A_2 = 2A/3 = 2 \times 10^{-2}/3\, m^2$
$v_p = 20\, m/sec$
Substituting the values:
$0.1 = (A/3) \times 20 + (2A/3) \times v_Q$
Multiply by $3/A$:
$0.1 \times (3/A) = 20 + 2 v_Q$
$0.3 / 10^{-2} = 20 + 2 v_Q$
$30 = 20 + 2 v_Q$
$10 = 2 v_Q$
$v_Q = 5\, m/sec$

(C) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of flow remains constant: $A_1 V_1 = A_2 V_2$.
Here,$d_1 = 3\, cm$ (radius $r_1 = 1.5\, cm$) and $d_2 = 6\, cm$ (radius $r_2 = 3.0\, cm$).
The velocity in the narrower pipe is $V_1 = 4\, m/s$.
Substituting the values into the equation: $\pi (r_1)^2 V_1 = \pi (r_2)^2 V_2$.
$\pi (1.5)^2 \times 4 = \pi (3.0)^2 \times V_2$.
$2.25 \times 4 = 9 \times V_2$.
$9 = 9 \times V_2$.
Therefore,$V_2 = 1\, m/s$.

(A) As water falls from a tap,it accelerates due to gravity,so its velocity $v$ increases as it moves downwards.
According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area $A$ and the velocity $v$ remains constant: $A \cdot v = \text{constant}$.
Since $v$ increases as the water falls,the cross-sectional area $A$ must decrease to keep the product $A \cdot v$ constant.
Therefore,the area of the water stream decreases.

(C) Using the equation of motion $v_2^2 = v_1^2 + 2gh$,where $v_1 = 0.04\, ms^{-1}$,$g = 10\, ms^{-2}$,and $h = 8 \times 10^{-1}\, m$:
$v_2^2 = (0.04)^2 + 2 \times 10 \times 0.8 = 0.0016 + 16 = 16.0016 \approx 16\, m^2s^{-2}$.
Thus,$v_2 \approx 4\, ms^{-1}$.
Using the equation of continuity $A_1v_1 = A_2v_2$,where $A = \pi (D/2)^2$:
$D_1^2 v_1 = D_2^2 v_2$
$(8 \times 10^{-3})^2 \times 0.04 = D_2^2 \times 4$
$64 \times 10^{-6} \times 0.04 = D_2^2 \times 4$
$D_2^2 = 16 \times 10^{-6} \times 0.04 = 0.64 \times 10^{-6}$
$D_2 = \sqrt{0.64 \times 10^{-6}} = 0.8 \times 10^{-3}\, m$.

(D) According to the equation of continuity for an ideal fluid,the product of the cross-sectional area and the velocity of the fluid remains constant: $A_1 v_1 = A_2 v_2$.
This implies that the velocity is inversely proportional to the area: $v \propto \frac{1}{A}$.
Since the area $A = \pi r^2 = \pi (d/2)^2$,we have $A \propto d^2$,where $d$ is the diameter.
Therefore,the ratio of the minimum velocity to the maximum velocity is given by the ratio of the minimum area to the maximum area:
$\frac{v_{\min}}{v_{\max}} = \frac{A_{\min}}{A_{\max}} = \left( \frac{d_{\min}}{d_{\max}} \right)^2$.
Substituting the given values $d_{\min} = 4.8 \; cm$ and $d_{\max} = 6.4 \; cm$:
$\frac{v_{\min}}{v_{\max}} = \left( \frac{4.8}{6.4} \right)^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}$.

(C) Area of cross-section of the spray pump,$A_{1} = 8 \; cm^{2} = 8 \times 10^{-4} \; m^{2}$.
Number of holes,$n = 40$. Diameter of each hole,$d = 1 \; mm = 1 \times 10^{-3} \; m$.
Radius of each hole,$r = d/2 = 0.5 \times 10^{-3} \; m$.
Area of cross-section of each hole,$a = \pi r^{2} = \pi (0.5 \times 10^{-3})^{2} \; m^{2}$.
Total area of $40$ holes,$A_{2} = n \times a = 40 \times \pi (0.5 \times 10^{-3})^{2} \; m^{2} \approx 31.416 \times 10^{-6} \; m^{2}$.
Speed of flow of liquid inside the tube,$V_{1} = 1.5 \; m/min = 1.5/60 \; m/s = 0.025 \; m/s$.
According to the equation of continuity,$A_{1} V_{1} = A_{2} V_{2}$.
$V_{2} = (A_{1} V_{1}) / A_{2} = (8 \times 10^{-4} \times 0.025) / (31.416 \times 10^{-6}) \approx 0.636 \; m/s$.
Rounding to the nearest option,the speed of ejection is $0.63 \; m/s$.

(N/A) The flow of a fluid is said to be steady if,at any given point,the velocity of each passing fluid particle remains constant in time.
This means that the velocity of any particle of the fluid remains the same while passing through a given point.
For example,let the velocity of each particle passing through point $P$ be $\overrightarrow{v_{P}}$,the velocity of each particle passing through point $Q$ be $\overrightarrow{v_{Q}}$,and the velocity of each particle passing through point $R$ be $\overrightarrow{v_{R}}$.
It is not necessary that $\overrightarrow{v_{P}} = \overrightarrow{v_{Q}} = \overrightarrow{v_{R}}$.
When a particle passes from one point to another,its velocity can change.
Steady flow is typically observed in low-velocity fluid motion. For example,the very slow motion of a water stream or slow-flowing wind.

(N/A) The path of motion of a fluid particle is called a line of flow.
$A$ streamline is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point.
In a steady flow,the path taken by a fluid particle is a streamline. Figure $(a)$ shows a typical trajectory of a fluid particle,which represents a streamline. The velocity of the particle at any point is in the direction of the tangent drawn at that point.
Streamline flow is defined as a flow in which streamlines can be clearly defined. This occurs when the velocity of the fluid at any given point remains constant over time.
In unsteady flow,the velocity at a point changes with time,so while the path of a particle (line of flow) can be defined,a fixed streamline cannot be defined.

$A$ tube of flow is defined as a bundle of streamlines that form a tubular region in a fluid flow.
Key characteristics of a tube of flow include:
$1$. Since the boundary of the tube of flow is formed by streamlines,no fluid can cross the boundary of the tube. The fluid enters at one end and leaves at the other.
$2$. The velocity of all fluid particles passing through any cross-section perpendicular to the direction of flow within the tube is considered to be uniform at a given instant.
$3$. When fluid flows through a physical pipe or tube at a very low speed (laminar flow),the entire pipe can be considered a single tube of flow.

(N/A) Characteristics of streamlines:
$(i)$ In a steady flow,streamlines can never intersect each other.
$(ii)$ The tangent drawn at any point on a streamline indicates the direction of the velocity of the fluid particle at that point.
$(iii)$ The velocity of each fluid particle passing through a specific point on a streamline is constant over time,but it may vary from point to point along the streamline.
$(iv)$ Widely spaced streamlines indicate a region of low speed,while closely spaced streamlines indicate a region of high speed.

(N/A) The path taken by a fluid particle in a flowing fluid is called its line of flow.
In a moving fluid,the velocity and direction of particles generally change over time. Therefore,particles passing through a single point may not follow the same path.
However,in steady flow,the velocity of any particle passing through a specific point remains constant over time.
Suppose the velocity of a particle passing through point $P$ is $\overrightarrow{V}_{P}$,and the velocities of particles passing through points $Q, R$,and $S$ are $\overrightarrow{V}_{Q}, \overrightarrow{V}_{R}$,and $\overrightarrow{V}_{S}$ respectively.
In steady flow,the path of a particle passing through point $P$ is shown in the figure. This steady path is known as a streamline.
In steady flow,the line of flow and the streamline coincide.
Streamline: $A$ streamline is a curve such that the tangent at any point on it gives the direction of the velocity of the fluid particle passing through that point.
Streamline flow: $A$ flow in which streamlines can be defined is called streamline flow.
In unsteady flow,a line of flow can be defined for a specific particle,but a streamline cannot be defined because the velocity at a point changes with time.

(N/A) In a steady flow,if we draw streamlines for every particle,they form a tube-like structure known as a tube of flow. No fluid particle enters or leaves this tube through its sides.
The figure shows a tube of flow with points $P$,$R$,and $Q$.
Let $v_{P}$,$v_{Q}$,and $v_{R}$ be the velocities of the fluid particles at points $P$,$Q$,and $R$ respectively,and let $A_{P}$,$A_{Q}$,and $A_{R}$ be the corresponding cross-sectional areas.
Since the flow is steady,the mass of fluid passing through any cross-section in a given time interval $\Delta t$ must be constant.
The mass of fluid $m_{P}$ passing through cross-section $A_{P}$ with velocity $v_{P}$ in time $\Delta t$ is given by:
$m_{P} = A_{P} v_{P} \Delta t \rho$,where $\rho$ is the density of the incompressible fluid.
Similarly,the masses of fluid passing through $Q$ and $R$ in the same time interval $\Delta t$ are:
$m_{Q} = A_{Q} v_{Q} \Delta t \rho$ and $m_{R} = A_{R} v_{R} \Delta t \rho$.
Since the fluid is incompressible and the flow is steady,the mass entering the tube must equal the mass leaving it. Therefore:
$m_{P} = m_{R} = m_{Q}$
Substituting the expressions:
$A_{P} v_{P} \Delta t \rho = A_{R} v_{R} \Delta t \rho = A_{Q} v_{Q} \Delta t \rho$
Canceling common terms $\Delta t$ and $\rho$:
$A_{P} v_{P} = A_{R} v_{R} = A_{Q} v_{Q}$
This is the equation of continuity for steady flow,which represents the law of conservation of mass for an incompressible fluid.
In general,$A v = \text{constant}$,which implies $v \propto \frac{1}{A}$.
The product $Av$ represents the volume flux or flow rate,which remains constant throughout the tube of flow. At narrower portions,the streamlines are closely spaced,indicating higher velocity,while at wider portions,the streamlines are widely spaced,indicating lower velocity.

(N/A) Critical speed: Steady flow is achieved at low flow speeds up to a limiting value,called critical speed.
Turbulent flow: In a fluid flow,if the velocity of the fluid changes erratically from point to point as well as from time to time,the flow is known as a turbulent flow. If the speed of the fluid becomes greater than the critical speed,the flow loses steadiness and becomes turbulent.
White water rapids: When a fast-flowing stream encounters rocks,small foamy whirlpool-like regions called 'white water rapids' are formed.

(N/A) Steady flow is defined as a type of fluid motion in which the velocity of the fluid particles at any given point in space remains constant with respect to time.
In other words,at any fixed point in the path of the fluid,the velocity vector does not change as time passes.
Mathematically,for a steady flow,the velocity $v$ at a point $(x, y, z)$ satisfies the condition $\frac{\partial v}{\partial t} = 0$.

(N/A) $1$. Line of Flow: $A$ line of flow is a curve whose tangent at any point is in the direction of the fluid velocity at that point. It represents the path taken by a fluid particle as it moves through space.
$2$. Streamline: $A$ streamline is a specific type of line of flow in a steady flow,where the velocity vector at every point along the line is tangent to the streamline. In steady flow,the pattern of streamlines remains constant over time,meaning fluid particles follow the same path.

(B) streamline is defined as a curve whose tangent at any point is in the direction of the velocity vector of the fluid particle at that point.
Therefore,the velocity of a fluid particle at any point on a streamline is always directed along the tangent to the streamline at that point.
This implies that no fluid particle can cross a streamline,as the velocity vector is always tangential to the path.

(B) No,two streamlines cannot cross each other.
If two streamlines were to intersect,a fluid particle arriving at the point of intersection would have two possible directions of motion simultaneously.
This would imply that the velocity vector of the fluid particle is not uniquely defined at that point.
Since the velocity of a fluid particle in a steady flow is uniquely determined by the flow field,such an intersection is physically impossible.

(B) tube of flow is defined as a bundle of streamlines that enclose a finite region of space through which a fluid flows.
Since streamlines cannot cross each other,the fluid particles inside a tube of flow remain inside it throughout their motion.
The boundary of the tube of flow is formed by the streamlines passing through the perimeter of a closed curve.
Therefore,the correct answer is $B$.

(A) According to the properties of streamlines in fluid dynamics,the spacing between streamlines is inversely proportional to the velocity of the fluid.
When streamlines are closely spaced,it indicates that the fluid is passing through a narrower region.
By the equation of continuity,$A_1v_1 = A_2v_2$,where $A$ is the cross-sectional area and $v$ is the velocity.
As the area $A$ decreases,the velocity $v$ must increase to maintain a constant flow rate.
Therefore,closely spaced streamlines represent regions of higher fluid velocity.

(B) According to the properties of streamlines,the spacing between streamlines is inversely proportional to the velocity of the fluid.
When streamlines are widely spaced,it indicates that the fluid is moving through a larger cross-sectional area,which results in a lower velocity.
Conversely,when streamlines are closely spaced,the fluid velocity is higher.
Therefore,if streamlines are widely spaced,the velocity of the fluid decreases.

(N/A) Laminar flow, also known as streamline flow, is a type of fluid flow in which the fluid particles move in smooth, well-defined paths or layers (streamlines) that do not cross each other.
In this type of flow, the velocity of the fluid at any given point remains constant over time.
Laminar flow typically occurs at low velocities and is characterized by a low Reynolds number $(Re < 2000)$.