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

(N/A) Steady flow is defined as a type of fluid flow in which the velocity of the fluid particles at any given point remains constant with respect to time.
An example of steady flow is the flow of water through a pipe with a constant cross-sectional area at a constant flow rate.
In this scenario,at any specific point inside the pipe,the velocity of the water particles passing through it does not change over time.

(B) 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,i.e.,$A_1v_1 = A_2v_2$.
This implies $v \propto \frac{1}{A}$.
Therefore,when the cross-sectional area $A$ decreases (in a narrow tube),the velocity $v$ of the fluid must increase to maintain a constant flow rate.

(A) The equation of continuity is derived from the principle of conservation of mass.
For an incompressible fluid flowing through a pipe of varying cross-section,the mass of fluid entering the pipe must equal the mass of fluid leaving the pipe in the same time interval.
Mathematically,this is expressed as $A_1 v_1 = A_2 v_2$,where $A$ is the cross-sectional area and $v$ is the velocity of the fluid.

(C) Since water is not stored anywhere,the volume flow rate of the Ganga must equal the sum of the volume flow rates of the Bhagirathi and the Alaknanda.
By the equation of continuity,we have:
$A_g v_g = A_b v_b + A_a v_a \quad \dots(i)$
Given the ratio of cross-sectional areas $A_b : A_a : A_g = 1 : 2 : 3$,we can write:
$A_b = x, A_a = 2x, A_g = 3x$
Given the ratio of speeds $v_b : v_a = 1 : 1.5 = 1 : \frac{3}{2}$,we can write:
$v_b = y, v_a = 1.5y = \frac{3}{2}y$
Substituting these values into equation $(i)$:
$3x \cdot v_g = x \cdot y + 2x \cdot \left(\frac{3}{2}y\right)$
$3x \cdot v_g = xy + 3xy = 4xy$
$v_g = \frac{4}{3}y$
Now,the ratio of the speed of the Ganga to that of the Alaknanda is:
$\frac{v_g}{v_a} = \frac{\frac{4}{3}y}{\frac{3}{2}y} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$

(B) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the fluid velocity remains constant at all points along the tube.
Mathematically,this is expressed as:
$A_1 v_1 = A_2 v_2$
To find the ratio of the speeds $\frac{v_1}{v_2}$,we rearrange the equation:
$\frac{v_1}{v_2} = \frac{A_2}{A_1}$
Thus,the correct option is $B$.

(A) According to the equation of continuity for an incompressible fluid in steady flow,the volume flow rate (discharge) remains constant throughout the pipe.
$A_P V_P = A_Q V_Q$
Where $A$ is the cross-sectional area and $V$ is the velocity of water.
Since $A = \frac{\pi d^2}{4}$,we have:
$\frac{\pi d_P^2}{4} V_P = \frac{\pi d_Q^2}{4} V_Q$
$d_P^2 V_P = d_Q^2 V_Q$
$V_P = \left( \frac{d_Q}{d_P} \right)^2 V_Q$
Given $d_P = 2 \times 10^{-2} \, m$ and $d_Q = 4 \times 10^{-2} \, m$:
$V_P = \left( \frac{4 \times 10^{-2}}{2 \times 10^{-2}} \right)^2 V_Q$
$V_P = (2)^2 V_Q$
$V_P = 4 V_Q$
Thus,the velocity of water in pipe $P$ is four times that of pipe $Q$.

(A) The Assertion is true because when water flows vertically upward,gravity acts against the motion,causing the velocity to decrease. By the equation of continuity $(A_1v_1 = A_2v_2)$,as velocity $(v)$ decreases,the cross-sectional area $(A)$ must increase,causing the stream to spread. Conversely,when flowing downward,gravity increases the velocity,causing the stream to narrow.
The Reason is also true; the equation of continuity is based on the principle of conservation of mass for an incompressible fluid,which states that the volume flow rate $(Q = Av)$ remains constant.
Since the change in the cross-sectional area of the stream is directly caused by the change in velocity due to gravity,and the relationship between area and velocity is governed by the constant volume flow rate,the Reason correctly explains the Assertion.

(C) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the fluid velocity remains constant at all points along the tube.
$A_1 V_1 = A_2 V_2$
Given:
$A_1 = 2.0\,cm^2 = 2.0 \times 10^2\,mm^2 = 200\,mm^2$
$V_1 = 4\,cm/s$
$A_2 = 10\,mm^2$
Substituting the values into the equation:
$200\,mm^2 \times 4\,cm/s = 10\,mm^2 \times V_2$
$800 = 10 \times V_2$
$V_2 = 80\,cm/s$
Therefore,the speed of the outgoing fluid is $80\,cm/s$.

(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 all points in the tube.
$A_1 v_1 = A_2 v_2$
Here,$A_1 = \pi r_1^2$ and $A_2 = \pi r_2^2$.
Given: $r_1 = 2 \ cm$,$r_2 = 1 \ cm$,and $v_1 = 2 \ m/s$.
Substituting the values:
$\pi (2)^2 \times 2 = \pi (1)^2 \times v_2$
$4 \times 2 = 1 \times v_2$
$v_2 = 8 \ m/s$.

(D) 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 flow: $A_1 V_1 = A_2 V_2$.
Here,the cross-sectional area of the pipe is $A = \pi r^2$.
At point $A$,the radius is $R$,so the area $A_A = \pi R^2$ and velocity $V_A = V$.
At point $B$,the radius is $\frac{R}{3}$,so the area $A_B = \pi \left(\frac{R}{3}\right)^2 = \frac{\pi R^2}{9}$.
Using the continuity equation: $(\pi R^2) \times V = \left(\frac{\pi R^2}{9}\right) \times V_B$.
Canceling $\pi R^2$ from both sides,we get: $V = \frac{V_B}{9}$.
Therefore,the velocity at point $B$ is $V_B = 9V$.

(C) The rate of flow (volume flow rate) is given by $Q = A \times v$,where $A$ is the cross-sectional area and $v$ is the velocity of the water.
Given,$Q = \pi \times 10^{-1} \,m^3/s$.
The radius of the pipe $r = 10 \,cm = 0.1 \,m$.
The area of the cross-section $A = \pi r^2 = \pi \times (0.1)^2 = \pi \times 0.01 \,m^2$.
Using the equation of continuity,$v = Q / A$.
$v = (\pi \times 10^{-1}) / (\pi \times 0.01) = 0.1 / 0.01 = 10 \,m/s$.
Therefore,the velocity of water is $10 \,m/s$.

(B) According to the equation of continuity,the mass flow rate of water remains constant throughout the system.
$A_1 v_1 = A_2 v_2$
Here,$A_1 = \pi R^2$ is the cross-sectional area of the pipe and $v_1 = v$ is the speed of water in the pipe.
The sprinkler has $n$ holes,each of radius $r$. The total cross-sectional area of the holes is $A_2 = n \pi r^2$.
Let $v'$ be the speed of water leaving the sprinkler.
Substituting these values into the continuity equation:
$\pi R^2 v = (n \pi r^2) v'$
Solving for $v'$:
$v' = \frac{\pi R^2 v}{n \pi r^2} = \frac{R^2 v}{n r^2}$

(A) According to the equation of continuity, the volume flow rate at the tank surface must equal the volume flow rate at the tap: $A_1 v_1 = A_2 v_2$.
Here, $A_1 = 750 \,cm^2 = 750 \times 10^{-4} \,m^2$ and $A_2 = 500 \,mm^2 = 500 \times 10^{-6} \,m^2$.
The speed of water at the tap is $v_2 = 30 \,cm/s = 0.3 \,m/s$.
The speed of the water surface falling is $v_1 = \frac{dh}{dt}$.
Substituting the values: $(750 \times 10^{-4}) \cdot \frac{dh}{dt} = (500 \times 10^{-6}) \times (0.3)$.
$\frac{dh}{dt} = \frac{500 \times 10^{-6} \times 0.3}{750 \times 10^{-4}} = \frac{150 \times 10^{-6}}{750 \times 10^{-4}} = 0.2 \times 10^{-2} \,m/s = 2 \times 10^{-3} \,m/s$.
Given $\frac{dh}{dt} = x \times 10^{-3} \,m/s$, we find $x = 2$.

(B) 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,$A_1$ is the area of the pipe,$V$ is the velocity in the pipe,$A_n$ is the area of the nozzle,and $V_1$ is the velocity at the nozzle.
$A_1 = \frac{\pi d^2}{4}$ and $A_n = \frac{\pi d_n^2}{4}$,where $d_n$ is the diameter of the nozzle.
Substituting these into the continuity equation:
$\frac{\pi d^2}{4} \times V = \frac{\pi d_n^2}{4} \times V_1$
$d^2 V = d_n^2 V_1$
$d_n^2 = d^2 \frac{V}{V_1}$
$d_n = d \sqrt{\frac{V}{V_1}}$

(A) In a streamline flow,the velocity of a fluid at a specific point is always constant over time. This is because each particle of the fluid follows a well-defined path,and there is no mixing or crossing of paths between particles. Therefore,the statement that the velocity at a given point is 'never constant' is incorrect.

(A) By definition,in a streamlined flow,the velocity of the fluid at any specific point remains constant over time. Although the velocity may vary from one point to another in the path of the fluid,at any fixed point,the velocity vector does not change with time.

(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 at any point along the pipe: $A_1 V_1 = A_2 V_2$.
Here,$A_1$ is the area of the pipe,$V$ is the velocity in the pipe,$A_2$ is the area of the nozzle,and $V_1$ is the velocity at the nozzle.
Since the area $A = \pi (d/2)^2$,we have $A \propto d^2$.
Substituting this into the continuity equation: $d^2 V = d_2^2 V_1$.
Solving for the nozzle diameter $d_2$: $d_2^2 = d^2 \frac{V}{V_1}$.
Taking the square root on both sides: $d_2 = d \sqrt{\frac{V}{V_1}}$.

(C) Given: Flow rate $Q = 0.314 \,m^3/s$ and radius $r = 10 \,cm = 0.1 \,m$.
Using the equation of continuity, $Q = A \times v$, where $A = \pi r^2$ is the cross-sectional area.
Substituting the values: $0.314 = \pi \times (0.1)^2 \times v$.
Taking $\pi \approx 3.14$, we get $0.314 = 3.14 \times 0.01 \times v$.
$0.314 = 0.0314 \times v$.
$v = \frac{0.314}{0.0314} = 10 \,m/s$.

(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 at all points along the pipe.
Therefore,$A_A v_A = A_B v_B$,where $A$ is the cross-sectional area and $v$ is the velocity.
The area of a circular cross-section is given by $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,where $d$ is the diameter.
Thus,$\frac{v_A}{v_B} = \frac{A_B}{A_A} = \frac{\frac{\pi (d_B)^2}{4}}{\frac{\pi (d_A)^2}{4}} = \left( \frac{d_B}{d_A} \right)^2$.
Given $d_A = 5 \ cm$ and $d_B = 10 \ cm$,we have:
$\frac{v_A}{v_B} = \left( \frac{10}{5} \right)^2 = (2)^2 = \frac{4}{1}$.
Hence,the ratio of the velocities of the fluid at $A$ and $B$ is $4:1$.

(D) According to the equation of continuity for an incompressible fluid in streamline flow, the volume flow rate $(Q = Av)$ remains constant at all cross-sections of the tube.
Therefore, the rate of liquid flow is the same at both points $M$ and $N$.

(B) According to the equation of continuity, the volume flow rate remains constant throughout the system.
$A_{1} v_{1} = A_{2} v_{2}$
Here, $A_{1} = 8 \,cm^{2} = 8 \times 10^{-4} \,m^{2}$.
The speed inside the tube is $v_{1} = 0.15 \,m \,min^{-1} = \frac{0.15}{60} \,m \,s^{-1} = 0.0025 \,m \,s^{-1}$.
The total area of the $40$ holes is $A_{2} = 40 \times 10^{-8} \,m^{2}$.
Substituting these values into the continuity equation:
$(8 \times 10^{-4}) \times (0.0025) = (40 \times 10^{-8}) \times v_{2}$
$v_{2} = \frac{8 \times 10^{-4} \times 0.0025}{40 \times 10^{-8}}$
$v_{2} = \frac{2 \times 10^{-6}}{40 \times 10^{-8}} = \frac{200}{40} = 5 \,m \,s^{-1}$.
Thus, the speed with which the liquid is ejected is $5 \,m \,s^{-1}$.

(B) Given: Radius of the tube $r_1 = 2 \,cm = 0.02 \,m$.
Radius of each hole $r_2 = 0.4 \,mm = 0.0004 \,m$.
Number of holes $n = 50$.
Speed of liquid inside the tube $v_1 = 0.04 \,ms^{-1}$.
According to the equation of continuity, the volume flow rate inside the tube must equal the total volume flow rate through all the holes:
$A_1 v_1 = n A_2 v_2$
Substituting the areas $\pi r_1^2 v_1 = n \pi r_2^2 v_2$
$r_1^2 v_1 = n r_2^2 v_2$
$(0.02)^2 \times 0.04 = 50 \times (0.0004)^2 \times v_2$
$4 \times 10^{-4} \times 0.04 = 50 \times 16 \times 10^{-8} \times v_2$
$1.6 \times 10^{-5} = 800 \times 10^{-8} \times v_2$
$1.6 \times 10^{-5} = 8 \times 10^{-6} \times v_2$
$v_2 = \frac{1.6 \times 10^{-5}}{8 \times 10^{-6}} = 0.2 \times 10 = 2 \,ms^{-1}$.

(A) Given: Diameter of the first pipe,$d_1 = 2 \,cm$. Radius $r_1 = 1 \,cm = 10^{-2} \,m$.
Diameter of the second pipe,$d_2 = 4 \,cm$. Radius $r_2 = 2 \,cm = 2 \times 10^{-2} \,m$.
According to the equation of continuity for an incompressible fluid,the volume flow rate remains constant: $A_1 v_1 = A_2 v_2$.
Here,$A = \pi r^2$ is the cross-sectional area.
Substituting the values: $\pi (r_1)^2 v_1 = \pi (r_2)^2 v_2$.
$(10^{-2})^2 v_1 = (2 \times 10^{-2})^2 v_2$.
$10^{-4} v_1 = 4 \times 10^{-4} v_2$.
$v_1 = 4 v_2$.
Therefore,the velocity of flow in the $2 \,cm$ pipe is $4$ times the velocity in the $4 \,cm$ pipe.

(B) Given: Internal diameter of the hose pipe, $d_1 = 4 \,cm$. Radius $r_1 = \frac{d_1}{2} = 2 \,cm = 2 \times 10^{-2} \,m$.
Speed of water through the hose pipe, $v_1 = 1 \,ms^{-1}$.
Speed of water through the nozzle, $v_2 = 4 \,ms^{-1}$.
Let the diameter of the nozzle be $d_2$ and its radius be $r_2$.
According to the equation of continuity for an incompressible fluid, $A_1 v_1 = A_2 v_2$.
Substituting the area $A = \pi r^2$, we get $\pi r_1^2 v_1 = \pi r_2^2 v_2$.
$r_2^2 = \frac{r_1^2 v_1}{v_2} = \frac{(2 \times 10^{-2} \,m)^2 \times 1 \,ms^{-1}}{4 \,ms^{-1}}$.
$r_2^2 = \frac{4 \times 10^{-4}}{4} \,m^2 = 10^{-4} \,m^2$.
Taking the square root, $r_2 = 10^{-2} \,m = 1 \,cm$.
The diameter of the nozzle is $d_2 = 2 r_2 = 2 \times 1 \,cm = 2 \,cm$.

(A) According to the equation of continuity,the volume flow rate of water remains constant.
For the first pipe,the area of cross-section is $A_1 = \pi (D/2)^2$ and the speed is $v$.
For the second pipe,the water flows out through $n$ holes,each of area $a = \pi (d/2)^2$. Let the speed of water leaving these holes be $v'$.
Equating the flow rates: $A_1 v = n \times a \times v'$.
Substituting the values: $\pi (D/2)^2 v = n \times \pi (d/2)^2 v'$.
Simplifying the equation: $(D^2/4) v = n (d^2/4) v'$.
Solving for $v'$: $v' = \frac{D^2 v}{n d^2}$.

(D) Given, the rate of flow of water, $Q = 12 \pi \text{ litre/minute}$.
Converting this to $SI$ units $(m^3/s)$:
$Q = \frac{12 \pi \times 10^{-3} \text{ m}^3}{60 \text{ s}} = 0.2 \pi \times 10^{-3} \text{ m}^3/s = 2 \pi \times 10^{-4} \text{ m}^3/s$.
The diameter of the pipe is $d = 2 \text{ cm} = 0.02 \text{ m}$, so the radius is $r = 0.01 \text{ m} = 10^{-2} \text{ m}$.
The cross-sectional area $A$ is given by $A = \pi r^2 = \pi (10^{-2})^2 = \pi \times 10^{-4} \text{ m}^2$.
Using the equation of continuity, $Q = A \times v$, where $v$ is the velocity of water:
$2 \pi \times 10^{-4} = (\pi \times 10^{-4}) \times v$.
Solving for $v$:
$v = \frac{2 \pi \times 10^{-4}}{\pi \times 10^{-4}} = 2 \text{ m/s}$.

(B) 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 flow:
$A_A v_A = A_B v_B$
Given that the pipe is cylindrical,the cross-sectional area $A = \pi r^2$.
At point $A$,the radius is $2R$,so $A_A = \pi (2R)^2 = 4\pi R^2$.
At point $B$,the radius is $R$,so $A_B = \pi R^2$.
The velocity at point $B$ is given as $v_B = 4v$.
Substituting these values into the continuity equation:
$4\pi R^2 \times v_A = \pi R^2 \times 4v$
Dividing both sides by $4\pi R^2$:
$v_A = v$

(A) According to the equation of continuity,the volume flow rate of water entering the first pipe must equal the total volume flow rate of water exiting through the $n$ holes in the second pipe.
Let $A_1$ be the cross-sectional area of the first pipe and $v$ be the speed of water in it.
$A_1 = \pi \left(\frac{D}{2}\right)^2$
Let $A_2$ be the area of each hole and $v'$ be the speed of water leaving each hole.
$A_2 = \pi \left(\frac{d}{2}\right)^2$
The total flow rate out is $n \times A_2 \times v'$.
Equating the flow rates: $A_1 v = n A_2 v'$
$\pi \left(\frac{D}{2}\right)^2 v = n \pi \left(\frac{d}{2}\right)^2 v'$
$\frac{D^2}{4} v = n \frac{d^2}{4} v'$
$v' = \frac{D^2 v}{n d^2}$

(C) The radius of each circular opening is $r_1 = 1 \,mm = 10^{-3} \,m$.
The cross-sectional area of each opening is $A_1 = \pi r_1^2 = \pi \times (10^{-3})^2 = \pi \times 10^{-6} \,m^2$.
The radius of the pipe is $r_2 = 2 \,cm = 0.02 \,m$.
The cross-sectional area of the pipe is $A_2 = \pi r_2^2 = \pi \times (0.02)^2 = 4\pi \times 10^{-4} \,m^2$.
The speed of water in the pipe is $v_2 = 25 \,cm/s = 0.25 \,m/s$.
Let $v_1$ be the speed of water as it exits the $n = 25$ openings.
According to the equation of continuity, the total flow rate is constant:
$n A_1 v_1 = A_2 v_2$
$25 \times (\pi \times 10^{-6}) \times v_1 = (4\pi \times 10^{-4}) \times 0.25$
$v_1 = \frac{4\pi \times 10^{-4} \times 0.25}{25 \times \pi \times 10^{-6}}$
$v_1 = \frac{10^{-4}}{25 \times 10^{-6}} = \frac{100}{25} = 4 \,m/s$.

(C) The power required to move a fluid through a pipe is given by the work done per unit time. For a horizontal pipe,the work done is primarily to overcome the kinetic energy of the fluid. The kinetic energy per unit time (power) is given by $P = \frac{1}{2} \dot{m} v^2$,where $\dot{m}$ is the mass flow rate and $v$ is the velocity of the fluid.
Since the mass flow rate $\dot{m} = \rho A v$ (where $\rho$ is density and $A$ is the cross-sectional area),we have $\dot{m} \propto v$. Thus,$P \propto \dot{m} v^2 \propto \dot{m} (\frac{\dot{m}}{\rho A})^2 \propto \dot{m}^3$.
If the rate of flow $\dot{m}$ is increased by a factor of $n$,the new power $P_1$ will be $P_1 \propto (n \dot{m})^3 = n^3 \dot{m}^3$.
Therefore,the ratio $\frac{P_1}{P_0} = \frac{n^3 \dot{m}^3}{\dot{m}^3} = n^3$.
Hence,the ratio $P_1: P_0 = n^3: 1$.

(B) According to the equation of continuity for an incompressible fluid,the volume flow rate remains constant: $A_1 v_1 = N A_2 v_2$.
Here,$A_1 = 30 \text{ cm}^2 = 3000 \text{ mm}^2$,$v_1 = 50 \text{ cm/s}$,$N = 10$,and $A_2 = 15 \text{ mm}^2$.
Substituting the values: $3000 \text{ mm}^2 \times 50 \text{ cm/s} = 10 \times 15 \text{ mm}^2 \times v_2$.
$150000 \text{ mm}^2 \cdot \text{cm/s} = 150 \text{ mm}^2 \times v_2$.
$v_2 = \frac{150000}{150} \text{ cm/s} = 1000 \text{ cm/s}$.
Converting to $SI$ units: $v_2 = 1000 \text{ cm/s} = 10 \text{ m/s}$.