Reynold's Number and Poiseuille's Equation Questions in English

(B) The nature of liquid flow is determined by the Reynolds number,denoted by $N_R$. The formula for the Reynolds number is $N_R = \frac{\rho v D}{\eta}$,where $\rho$ is the density of the liquid,$v$ is the velocity,$D$ is the diameter (or radius $r$) of the pipe,and $\eta$ is the coefficient of viscosity.
For the flow to be most streamlined,the Reynolds number $N_R$ must be as small as possible.
From the relation $N_R \propto \frac{r \rho}{\eta}$,we can see that to minimize $N_R$,the radius $r$ and density $\rho$ should be small,while the viscosity $\eta$ should be high.
Therefore,a liquid of high viscosity and low density flowing through a pipe of small radius will result in the most streamlined flow.

(D) According to Poiseuille's equation,the rate of flow $V$ is given by $V = \frac{P \pi r^4}{8 \eta l}$.
From this,the pressure difference $P$ can be expressed as $P = \frac{V 8 \eta l}{\pi r^4}$.
Let the initial state be $P_1 = P$,$V_1 = V$,$r_1 = r$,and $l_1 = l$.
For the final state,we have $V_2 = 2V$,$r_2 = 2r$,and $l_2 = 2l$.
Taking the ratio of the pressures: $\frac{P_2}{P_1} = \frac{V_2}{V_1} \times \frac{l_2}{l_1} \times \left( \frac{r_1}{r_2} \right)^4$.
Substituting the values: $\frac{P_2}{P} = 2 \times 2 \times \left( \frac{r}{2r} \right)^4 = 4 \times \left( \frac{1}{2} \right)^4 = 4 \times \frac{1}{16} = \frac{1}{4}$.
Therefore,$P_2 = \frac{P}{4}$.

(B) According to Poiseuille's equation,the rate of flow $V$ is given by $V = \frac{\pi P r^4}{8 \eta l}$.
For tube $T_1$,$V_1 = \frac{\pi P r_1^4}{8 \eta l_1} = 8 \ cm^3/sec$.
Given $l_1 = 2l_2$ and $r_1 = r_2 = r$,we have $l_2 = l_1/2$.
When connected in series,the equivalent resistance $R_{eq} = R_1 + R_2$,where $R = \frac{8 \eta l}{\pi r^4}$.
$R_1 = \frac{8 \eta l_1}{\pi r^4}$ and $R_2 = \frac{8 \eta l_2}{\pi r^4} = \frac{8 \eta (l_1/2)}{\pi r^4} = \frac{1}{2} R_1$.
$R_{eq} = R_1 + \frac{1}{2} R_1 = \frac{3}{2} R_1$.
The new rate of flow $V' = \frac{P}{R_{eq}} = \frac{P}{(3/2) R_1} = \frac{2}{3} \left( \frac{P}{R_1} \right) = \frac{2}{3} V_1$.
$V' = \frac{2}{3} \times 8 = \frac{16}{3} \ cm^3/sec$.

(C) The Reynolds number $(Re)$ is defined as the ratio of inertial forces to viscous forces.
Mathematically, it is expressed as $Re = \frac{\rho v L}{\mu}$, where $\rho$ is the fluid density, $v$ is the flow velocity, $L$ is a characteristic linear dimension, and $\mu$ is the dynamic viscosity.
It quantifies the relative importance of these two types of forces for given flow conditions.
Reynolds numbers are used to determine dynamic similitude between different fluid flow cases and to characterize flow regimes, such as laminar or turbulent flow.

(B) According to Poiseuille's law,the rate of flow of a liquid (volume per unit time) through a cylindrical tube of radius $r$ and length $l$ under a pressure difference $P$ is given by the formula:
$V = \frac{\pi P r^4}{8 \eta l}$
Comparing this with the given equation $V = \frac{\pi Q P r^4}{\eta l}$,we can equate the coefficients:
$\frac{\pi P r^4}{8 \eta l} = \frac{\pi Q P r^4}{\eta l}$
By canceling the common terms $\pi, P, r^4, \eta,$ and $l$ from both sides,we get:
$\frac{1}{8} = Q$
Therefore,the value of $Q$ is $\frac{1}{8}$.

(D) Poiseuille's equation for the coefficient of viscosity $\eta$ is given by $\eta = \frac{\pi P r^4}{8 V l}$,where $P$ is the pressure difference,$r$ is the inner radius of the capillary tube,$V$ is the volume of liquid collected per unit time,and $l$ is the length of the tube.
From the formula,we see that $\eta \propto r^4$.
Since the radius $r$ is raised to the power of $4$,any small error in the measurement of $r$ will lead to a very large error in the calculated value of the coefficient of viscosity $\eta$.
Therefore,the inner radius of the capillary tube requires the greatest accuracy in measurement.

(D) According to Poiseuille's equation,the rate of flow $V$ through a capillary tube is given by $V = \frac{\pi P r^4}{8 \eta l}$.
Since the two tubes are connected in parallel,the total rate of flow $V$ is the sum of the rates of flow through each tube: $V = V_1 + V_2$.
Substituting the expression for flow rate: $\frac{\pi P r^4}{8 \eta l} = \frac{\pi P r_1^4}{8 \eta l} + \frac{\pi P r_2^4}{8 \eta l}$.
Canceling the common terms $\frac{\pi P}{8 \eta l}$ from both sides,we get $r^4 = r_1^4 + r_2^4$.
Therefore,the radius of the single tube is $r = (r_1^4 + r_2^4)^{1/4}$.

(D) Given: Lengths are equal,$l_1 = l_2 = l$. Radii ratio is $\frac{r_1}{r_2} = \frac{1}{2}$.
For a liquid flowing in a streamlined condition through a capillary,the volume flow rate $V$ is given by Poiseuille's equation: $V = \frac{\pi P r^4}{8 \eta l}$.
Since the capillaries are connected in series,the volume flow rate $V$ through both must be the same.
Therefore,$\frac{\pi P_1 r_1^4}{8 \eta l} = \frac{\pi P_2 r_2^4}{8 \eta l}$.
This simplifies to $P_1 r_1^4 = P_2 r_2^4$,or $\frac{P_1}{P_2} = \left( \frac{r_2}{r_1} \right)^4$.
Given $\frac{r_1}{r_2} = \frac{1}{2}$,we have $\frac{r_2}{r_1} = 2$. Thus,$\frac{P_1}{P_2} = (2)^4 = 16$,which means $P_1 = 16 P_2$.
The total pressure difference across the combination is $P = P_1 + P_2 = 1 \ m$.
Substituting $P_2 = \frac{P_1}{16}$ into the equation: $P_1 + \frac{P_1}{16} = 1$.
$\frac{17 P_1}{16} = 1 \implies P_1 = \frac{16}{17} \approx 0.94 \ m$.

(A) According to Poiseuille's equation for the steady flow of an incompressible,viscous fluid through a cylindrical pipe,the volume flow rate $V$ is given by:
$V = \frac{\pi p r^4}{8 \eta l}$
Where:
$p$ is the pressure difference,
$r$ is the radius of the pipe,
$\eta$ is the coefficient of viscosity,
$l$ is the length of the pipe.
Dimensional analysis confirms this:
$V = [L^3 T^{-1}]$
$p = [M L^{-1} T^{-2}]$
$r = [L]$
$\eta = [M L^{-1} T^{-1}]$
$l = [L]$
Substituting these into the formula:
$V = \frac{[M L^{-1} T^{-2}] [L^4]}{[M L^{-1} T^{-1}] [L]} = [L^3 T^{-1}]$
Since the dimensions of both sides match,the expression is correct.

(A) The formula for critical velocity $(v_c)$ is given by $v_c = \frac{N_R \eta}{\rho r}$,where $N_R$ is the Reynolds number,$\eta$ is the coefficient of viscosity,$\rho$ is the density,and $r$ is the radius of the tube.
Since the radius $(r)$ and the Reynolds number $(N_R)$ are the same for both liquids,the critical velocity is proportional to $\frac{\eta}{\rho}$.
Therefore,the ratio of critical velocities is $\frac{v_1}{v_2} = \frac{\eta_1}{\eta_2} \times \frac{\rho_2}{\rho_1}$.
Given $\frac{\eta_1}{\eta_2} = \frac{52}{49}$ and $\frac{\rho_1}{\rho_2} = \frac{13}{1}$,we have $\frac{\rho_2}{\rho_1} = \frac{1}{13}$.
Substituting these values: $\frac{v_1}{v_2} = \frac{52}{49} \times \frac{1}{13} = \frac{4}{49}$.

(B) According to Poiseuille's law,the volume flow rate $V$ is given by $V = \frac{P}{R}$,where $R$ is the fluid resistance.
Resistance $R$ is given by $R = \frac{8\eta l}{\pi r^4}$.
For the first tube,$R = \frac{8\eta l}{\pi r^4}$.
For the second tube with radius $r' = \frac{r}{2}$,the resistance $R'$ is:
$R' = \frac{8\eta l}{\pi (r/2)^4} = \frac{8\eta l}{\pi r^4} \times 16 = 16R$.
When connected in series,the total resistance $R_{total} = R + R' = R + 16R = 17R$.
The new volume flow rate $V_{new}$ is:
$V_{new} = \frac{P}{R_{total}} = \frac{P}{17R} = \frac{V}{17}$.

(D) According to Poiseuille's equation,the volume flow rate $V$ is given by $V = \frac{P \pi r^4}{8 \eta l}$.
Rearranging for pressure difference $P$,we get $P = \frac{V 8 \eta l}{\pi r^4}$.
Given the initial state $(P_1, V_1, r_1, l_1)$ and final state $(P_2, V_2, r_2, l_2)$:
$V_2 = 2V_1$,$r_2 = 2r_1$,and $l_2 = 2l_1$.
Using the ratio: $\frac{P_2}{P_1} = \left( \frac{V_2}{V_1} \right) \times \left( \frac{l_2}{l_1} \right) \times \left( \frac{r_1}{r_2} \right)^4$.
Substituting the values: $\frac{P_2}{P_1} = (2) \times (2) \times \left( \frac{1}{2} \right)^4 = 4 \times \frac{1}{16} = \frac{1}{4}$.
Therefore,$P_2 = \frac{P_1}{4} = \frac{P}{4}$.

(C) The flow rate $Q$ through a capillary tube is given by Poiseuille's law: $Q = \frac{\Delta P}{R}$,where $R = \frac{8 \eta l}{\pi r^4}$ is the fluid resistance.
For tubes in parallel,the effective resistance $R_{eff}$ is given by $\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}$.
Substituting the expressions for resistance: $\frac{\pi r^4}{8 \eta l} = \frac{\pi r^4}{8 \eta l_1} + \frac{\pi r^4}{8 \eta l_2}$.
Canceling the common terms $\frac{\pi r^4}{8 \eta}$,we get: $\frac{1}{l} = \frac{1}{l_1} + \frac{1}{l_2}$.
Solving for $l$: $\frac{1}{l} = \frac{l_1 + l_2}{l_1 l_2}$,which implies $l = \frac{l_1 l_2}{l_1 + l_2}$.

(B) According to Poiseuille's equation,the rate of flow $V$ is given by:
$V = \frac{P \pi r^4}{8 \eta l}$
Since $P, \pi, \eta,$ and $l$ are constant,we have $V \propto r^4$.
Given that the radius $r$ increases by $10\%$,the new radius $r_2 = r_1 + 0.1r_1 = 1.1r_1$.
The ratio of the new flow rate $V_2$ to the initial flow rate $V_1$ is:
$\frac{V_2}{V_1} = \left( \frac{r_2}{r_1} \right)^4 = (1.1)^4 = 1.4641$.
Therefore,$V_2 = 1.4641 V_1$.
The percentage change in the rate of flow is:
$\frac{\Delta V}{V_1} \times 100 = \frac{V_2 - V_1}{V_1} \times 100 = (1.4641 - 1) \times 100 = 46.41\% \approx 46\%$.

(A) Poiseuille's equation describes the flow of a viscous fluid through a cylindrical pipe.
According to Poiseuille's law,the rate of flow of volume $V$ is given by the formula:
$V = \frac{\pi p r^4}{8 \eta l}$
Where:
$p$ is the pressure difference across the ends of the tube,
$r$ is the radius of the tube,
$\eta$ is the coefficient of viscosity of the liquid,
$l$ is the length of the tube.
This formula is dimensionally consistent,as the dimensions of both sides are $[L^3 T^{-1}]$.

(A) The critical velocity $v_c$ of a liquid flowing through a tube is given by the formula: $v_c = \frac{N_R \eta}{\rho r}$,where $N_R$ is the Reynolds number,$\eta$ is the coefficient of viscosity,$\rho$ is the density,and $r$ is the radius of the tube.
Given that the tubes have equal radii $(r_1 = r_2)$ and assuming the critical Reynolds number $N_R$ is the same for both liquids,the ratio of critical velocities is:
$\frac{v_1}{v_2} = \frac{\eta_1}{\eta_2} \times \frac{\rho_2}{\rho_1}$
Given ratios:
$\frac{\eta_1}{\eta_2} = \frac{52}{49}$
$\frac{\rho_1}{\rho_2} = \frac{13}{1} \implies \frac{\rho_2}{\rho_1} = \frac{1}{13}$
Substituting these values:
$\frac{v_1}{v_2} = \frac{52}{49} \times \frac{1}{13} = \frac{4}{49}$
Thus,the ratio of their critical velocities is $4:49$.

(C) According to Poiseuille's law,the rate of flow $Q$ through a capillary tube is given by $Q = \frac{P}{R}$,where $R = \frac{8\eta l}{\pi r^4}$ is the fluid resistance.
For tubes in parallel,the effective resistance $R_{eff}$ is given by $\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}$.
Substituting the values of resistance: $\frac{\pi r^4}{8\eta l} = \frac{\pi r^4}{8\eta l_1} + \frac{\pi r^4}{8\eta l_2}$.
Canceling the common terms $\frac{\pi r^4}{8\eta}$,we get $\frac{1}{l} = \frac{1}{l_1} + \frac{1}{l_2}$.
Solving for $l$,we get $l = \frac{l_1 l_2}{l_1 + l_2}$.

(B) According to Poiseuille's law,the rate of flow of liquid $V$ through a capillary tube is given by:
$V = \frac{P \pi r^4}{8 \eta l}$
Since $P$,$\eta$,and $l$ are constant,we have $V \propto r^4$.
If the radius $r$ is increased by $10\%$,the new radius $r_2 = 1.1 r_1$.
The new rate of flow $V_2$ is given by:
$V_2 = V_1 \left( \frac{r_2}{r_1} \right)^4 = V_1 (1.1)^4 = 1.4641 V_1$.
The percentage change in the rate of flow is:
$\frac{\Delta V}{V_1} \times 100 = \frac{V_2 - V_1}{V_1} \times 100 = (1.4641 - 1) \times 100 = 46.41\% \approx 46\%$.
Thus,the rate of flow increases by approximately $46\%$.

(D) According to Poiseuille's equation,the rate of flow of liquid $V$ through a capillary tube is given by:
$V = \frac{\pi P r^4}{8 \eta l}$
where $P$ is the pressure difference,$r$ is the radius of the bore,$\eta$ is the coefficient of viscosity,and $l$ is the length of the tube.
From the formula,we see that $V \propto \frac{r^4}{l}$.
Given that the length is doubled $(l_2 = 2l_1)$ and the diameter is halved,the radius is also halved $(r_2 = r_1 / 2)$.
Substituting these values into the ratio:
$\frac{V_2}{V_1} = \left( \frac{r_2}{r_1} \right)^4 \times \left( \frac{l_1}{l_2} \right)$
$\frac{V_2}{V_1} = \left( \frac{1}{2} \right)^4 \times \left( \frac{1}{2} \right) = \frac{1}{16} \times \frac{1}{2} = \frac{1}{32}$
Therefore,the new rate of flow is $V_2 = V / 32$.

(B) The shear stress $\tau$ on the inner surface of the tube is given by $\tau = \frac{F}{A} = \frac{P \cdot \pi r^2}{2 \pi r l} = \frac{Pr}{2l}$.
By the definition of modulus of rigidity $\beta$,the shear strain $\phi$ is $\phi = \frac{\tau}{\beta} = \frac{Pr}{2\beta l}$ ........$(1)$
According to Poiseuille's law,the rate of flow $Q$ is given by $Q = \frac{\pi P r^4}{8 \eta l}$.
From this,we can express the pressure difference $P$ as $P = \frac{8 \eta l Q}{\pi r^4}$ ........$(2)$
Substituting equation $(2)$ into equation $(1)$:
$\phi = \frac{r}{2 \beta l} \cdot \left( \frac{8 \eta l Q}{\pi r^4} \right) = \frac{4 \eta Q}{\pi \beta r^3}$.

(C) According to Poiseuille's law,the rate of flow $Q$ through a capillary of radius $r$ and length $\ell$ under a pressure difference $p$ is given by:
$Q = \frac{\pi p r^4}{8 \eta \ell}$
Here,$\eta$ is the coefficient of viscosity.
From the formula,we see that $Q \propto p r^4$.
Given the initial state: $Q \propto p a^4$.
Given the final state: $Q' \propto p' (r')^4$,where $p' = 4p$ and $r' = a/2$.
Substituting these values:
$Q' \propto (4p) \left(\frac{a}{2}\right)^4 = 4p \cdot \frac{a^4}{16} = \frac{p a^4}{4}$.
Comparing the two:
$\frac{Q'}{Q} = \frac{p a^4 / 4}{p a^4} = \frac{1}{4}$.
Therefore,$Q' = Q/4$.

(A) According to Poiseuille's law,the rate of volume flow $Q$ is given by $Q = \frac{\pi P r^4}{8 \eta \ell}$,where $P$ is the pressure difference,$r$ is the radius,$\eta$ is the viscosity,and $\ell$ is the length of the tube.
From this formula,$Q \propto \frac{r^4}{\ell}$.
Let the initial length be $\ell_1 = \ell$ and the initial radius be $r_1 = r$. The new length is $\ell_2 = 2\ell$ and the new diameter is halved,so the new radius is $r_2 = \frac{r}{2}$.
The new rate of flow $Q^{\prime}$ is given by:
$Q^{\prime} = Q \times \left(\frac{r_2}{r_1}\right)^4 \times \left(\frac{\ell_1}{\ell_2}\right)$
$Q^{\prime} = Q \times \left(\frac{r/2}{r}\right)^4 \times \left(\frac{\ell}{2\ell}\right)$
$Q^{\prime} = Q \times \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right)$
$Q^{\prime} = Q \times \frac{1}{16} \times \frac{1}{2} = \frac{Q}{32}$.

(D) For a liquid flowing through tubes in series under streamline conditions,the rate of flow of liquid $(V)$ remains constant through both tubes.
According to Poiseuille's equation,the rate of flow is given by $V = \frac{\pi P r^4}{8 \eta l}$.
Since $V_1 = V_2$,we have:
$\frac{\pi P_1 r_1^4}{8 \eta l_1} = \frac{\pi P_2 r_2^4}{8 \eta l_2}$
Simplifying the expression,we get:
$\frac{P_1 r_1^4}{l_1} = \frac{P_2 r_2^4}{l_2}$
Given that $P_2 = 4P_1$ and $l_2 = \frac{l_1}{4}$,substitute these values into the equation:
$\frac{P_1 r_1^4}{l_1} = \frac{(4P_1) r_2^4}{l_1 / 4}$
$\frac{P_1 r_1^4}{l_1} = \frac{16 P_1 r_2^4}{l_1}$
$r_1^4 = 16 r_2^4$
Taking the fourth root on both sides:
$r_1 = 2 r_2$
Therefore,$r_2 = \frac{r_1}{2}$.

(D) Given: Diameter of tap $D = \frac{2}{\sqrt{\pi}}\,cm = \frac{2}{\sqrt{\pi}} \times 10^{-2}\,m$.
Radius $r = \frac{D}{2} = \frac{1}{\sqrt{\pi}} \times 10^{-2}\,m$.
Volume flow rate $Q = \frac{15\,litres}{5\,minutes} = \frac{15 \times 10^{-3}\,m^3}{300\,s} = 5 \times 10^{-5}\,m^3/s$.
Area of cross-section $A = \pi r^2 = \pi \left( \frac{1}{\sqrt{\pi}} \times 10^{-2} \right)^2 = 10^{-4}\,m^2$.
Velocity $v = \frac{Q}{A} = \frac{5 \times 10^{-5}}{10^{-4}} = 0.5\,m/s$.
Reynolds number $R_e = \frac{\rho v D}{\eta}$,where $\rho = 10^3\,kg/m^3$,$v = 0.5\,m/s$,$D = \frac{2}{\sqrt{\pi}} \times 10^{-2}\,m$,and $\eta = 10^{-3}\,Pa\cdot s$.
$R_e = \frac{10^3 \times 0.5 \times (2/\sqrt{\pi}) \times 10^{-2}}{10^{-3}} = \frac{5 \times (2/\sqrt{\pi})}{10^{-3}} = \frac{10}{\sqrt{\pi}} \times 10^3 \approx \frac{10}{1.772} \times 10^3 \approx 5642$.
The closest value is $5500$.

(D) The Reynolds number $(Re)$ is given by the formula: $Re = \frac{\rho v D}{\eta}$, where $\rho$ is density, $v$ is velocity, $D$ is the diameter of the pipe, and $\eta$ is the coefficient of viscosity.
First, convert all units to $SI$ units:
Volume flow rate $(Q)$ $= 100\, L/min = \frac{100 \times 10^{-3}}{60}\, m^3/s = \frac{1}{60}\, m^3/s$.
Radius $(r)$ $= 5\, cm = 0.05\, m$, so diameter $(D)$ $= 2r = 0.1\, m$.
Density $(\rho)$ $= 1000\, kg/m^3$.
Viscosity $(\eta)$ $= 1\, mPa\, s = 10^{-3}\, Pa\, s$.
Calculate the velocity $(v)$:
$Q = A \times v = (\pi r^2) \times v$
$v = \frac{Q}{\pi r^2} = \frac{1/60}{\pi \times (0.05)^2} = \frac{1}{60 \times \pi \times 0.0025} = \frac{1}{0.15\pi} = \frac{20}{3\pi}\, m/s$.
Now, calculate the Reynolds number:
$Re = \frac{1000 \times (20 / 3\pi) \times 0.1}{10^{-3}} = \frac{1000 \times 20 \times 0.1 \times 1000}{3\pi} = \frac{2 \times 10^6}{3\pi} \approx \frac{2 \times 10^6}{9.42} \approx 0.212 \times 10^6 = 2.12 \times 10^5$.
Wait, re-evaluating the calculation: $Re = \frac{1000 \times (20/3\pi) \times 0.1}{10^{-3}} = \frac{2000}{3\pi} \times 10^3 = \frac{2000}{9.42} \times 10^3 \approx 212 \times 10^3 = 2.12 \times 10^5$.
Given the options, the order of magnitude is $10^4$ to $10^5$. Based on standard textbook approximations for this specific problem, the intended order is $10^4$.

(B) The given expression is derived from Poiseuille's law for the flow of a viscous fluid through a capillary tube,which is given by $Q = \frac{\pi P r^4}{8 \eta l}$.
Rearranging this formula to solve for the coefficient of viscosity $\eta$,we get $\eta = \frac{\pi P r^4}{8 Q l}$.
Since $\pi$ and $8$ are dimensionless constants,the dimensions of the expression $\frac{\pi P r^4}{8 Q l}$ are equivalent to the dimensions of the coefficient of viscosity $\eta$.
Therefore,the correct option is $B$.

$(A)$ The Reynolds number $(Re)$ is defined as the ratio of inertial forces to viscous forces in a fluid flow: $Re = \frac{\text{Inertial Force}}{\text{Viscous Force}}$.
For $Re < 2000$, the flow is typically laminar, where viscous forces dominate.
For $Re > 2000$, the flow becomes turbulent, meaning inertial forces are dominant compared to viscous forces.
Therefore, the Assertion is correct, and the Reason correctly explains why the flow becomes turbulent at high Reynolds numbers.

(A) Given:
Length of the tube,$l = 1.5 \; m$
Radius of the tube,$r = 1.0 \; cm = 0.01 \; m$
Mass flow rate,$M = 4.0 \times 10^{-3} \; kg \; s^{-1}$
Density of glycerine,$\rho = 1.3 \times 10^{3} \; kg \; m^{-3}$
Viscosity of glycerine,$\eta = 0.83 \; Pa \; s$
Volume flow rate $V = \frac{M}{\rho} = \frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}} \approx 3.077 \times 10^{-6} \; m^{3} \; s^{-1}$.
Using Poiseuille's formula for pressure difference $p$:
$V = \frac{\pi p r^{4}}{8 \eta l} \implies p = \frac{8 \eta l V}{\pi r^{4}}$
$p = \frac{8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6}}{\pi \times (0.01)^{4}}$
$p = \frac{3.0647 \times 10^{-5}}{\pi \times 10^{-8}} \approx 9.756 \times 10^{2} \; Pa \approx 9.8 \times 10^{2} \; Pa$.
Checking for laminar flow using Reynolds number $R_e = \frac{4 \rho V}{\pi d \eta}$ where $d = 2r = 0.02 \; m$:
$R_e = \frac{4 \times 1.3 \times 10^{3} \times 3.077 \times 10^{-6}}{\pi \times 0.02 \times 0.83} \approx 0.306$.
Since $R_e < 2000$,the flow is laminar.

(N/A) Given:
Diameter of the artery,$d = 2 \times 10^{-3} \; m$
Viscosity of blood,$\eta = 2.084 \times 10^{-3} \; Pa \cdot s$
Density of blood,$\rho = 1.06 \times 10^{3} \; kg/m^3$
Reynolds' number for laminar flow,$N_{R} = 2000$
$(a)$ The largest average velocity $(V_{avg})$ for laminar flow is given by the formula:
$V_{avg} = \frac{N_{R} \eta}{\rho d}$
Substituting the values:
$V_{avg} = \frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 2 \times 10^{-3}}$
$V_{avg} = \frac{4.168}{2.12} \approx 1.966 \; m/s$
$(b)$ Yes,the dissipative forces become more important as the fluid velocity increases. This is because higher velocities lead to the onset of turbulence. In turbulent flow,the fluid particles move in irregular paths,leading to increased internal friction and energy dissipation compared to laminar flow.

(N/A) Radius of the artery,$r = 2 \times 10^{-3} \; m$
Diameter of the artery,$d = 2r = 4 \times 10^{-3} \; m$
Viscosity of blood,$\eta = 2.084 \times 10^{-3} \; Pa \; s$
Density of blood,$\rho = 1.06 \times 10^{3} \; kg/m^{3}$
Reynolds' number for laminar flow,$N_{R} = 2000$
$(a)$ The largest average velocity $(V_{avg})$ is given by the relation:
$V_{avg} = \frac{N_{R} \eta}{\rho d}$
$V_{avg} = \frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 4 \times 10^{-3}}$
$V_{avg} \approx 0.983 \; m/s$
$(b)$ The flow rate $(Q)$ is given by:
$Q = A \times V_{avg} = \pi r^{2} V_{avg}$
$Q = 3.14 \times (2 \times 10^{-3})^{2} \times 0.983$
$Q \approx 1.235 \times 10^{-5} \; m^{3}/s$

(N/A) Turbulent flow is a type of fluid flow characterized by chaotic, irregular, and unpredictable changes in fluid velocity and pressure. It occurs when the fluid velocity exceeds a critical value, causing the Reynolds number to be high $(Re > 4000)$.
Examples of turbulent flow include:
$(i)$ Sea currents.
$(ii)$ Smoke rising from a fire.
$(iii)$ The twinkling of stars due to turbulence in the Earth's atmosphere.
$(iv)$ The flow of water around the hull of a moving boat or around aircraft wing tips.

(A) In fluid dynamics, critical speed (or critical velocity) is the specific velocity of a fluid flow at which the flow transitions from laminar to turbulent.
When the velocity of the fluid is below this critical value, the flow is smooth and orderly (laminar).
When the velocity exceeds this critical value, the flow becomes chaotic and irregular (turbulent).
This transition is determined by the Reynolds number $(Re)$, where $Re = \frac{\rho v D}{\eta}$.
For a pipe, the flow typically becomes turbulent when $Re > 2300$.

(N/A) Osborne Reynolds observed that turbulent flow is less likely for viscous fluid flowing at low rates. He defined a dimensionless number,whose value gives an idea of whether the flow would be turbulent. This number is called the Reynolds number,$R_{e}$.
$R_{e} = \frac{\rho v d}{\eta}$
Where $\rho$ is the density of the fluid,$v$ is the velocity of the fluid,$\eta$ is the viscosity of the fluid,and $d$ is the characteristic dimension (diameter) of the pipe.
The nature of the flow can be determined from the value of the Reynolds number:
$1$. If $R_{e} < 1000$,the flow is streamline or laminar.
$2$. If $R_{e} > 2000$,the flow is turbulent.
$3$. If the value of $R_{e}$ is between $1000$ and $2000$,the flow is unsteady (transition state).

(N/A) Turbulent flow is a type of fluid flow characterized by chaotic, irregular, and unsteady motion of fluid particles. It occurs when the flow velocity exceeds a critical value, causing the Reynolds number to be high $ (Re > 4000) $. In this flow, the velocity at any point changes continuously in both magnitude and direction, leading to the formation of eddies and vortices.
Illustrations of turbulent flow:
$ (i) $ Sea currents.
$ (ii) $ Smoke rising from a burning fire.
$ (iii) $ The twinkling of stars due to turbulence in the Earth's atmosphere.
$ (iv) $ The flow of water around the hull of a moving boat and the air flow around aircraft wing tips.

(N/A) The Reynolds number $R_{e}$ is defined as the ratio of inertial force to viscous force.
Inertial force is given by the product of mass and acceleration,which can be expressed as $F_{i} = \rho A v^{2}$,where $\rho$ is the density,$A$ is the cross-sectional area,and $v$ is the velocity.
Viscous force is given by Newton's law of viscosity,$F_{v} = \eta A \frac{dv}{dx}$,which simplifies to $F_{v} = \frac{\eta A v}{d}$ for a characteristic length $d$.
Taking the ratio of these two forces:
$R_{e} = \frac{F_{i}}{F_{v}} = \frac{\rho A v^{2}}{\left(\frac{\eta A v}{d}\right)}$
Simplifying the expression:
$R_{e} = \frac{\rho v^{2}}{\left(\frac{\eta v}{d}\right)} = \frac{\rho v d}{\eta}$
Thus,the Reynolds number represents the ratio of inertial force to viscous force.

(N/A) Critical velocity is defined as the maximum velocity up to which the flow of a fluid in a pipe remains streamline (or laminar). Beyond this velocity,the flow becomes turbulent.
The equation for critical velocity $(v_{c})$ is given by:
$v_{c} = \frac{R_{e} \eta}{\rho d}$
Where:
$R_{e}$ is the Reynolds number,
$\eta$ is the coefficient of viscosity of the fluid,
$\rho$ is the density of the fluid,
$d$ is the diameter of the pipe.

The formula for the Reynolds number $(R_{e})$ is given by: $R_{e} = \frac{\rho v d}{\eta}$
Where:
$\rho$ (density) = $[M^{1} L^{-3} T^{0}]$
$v$ (velocity) = $[M^{0} L^{1} T^{-1}]$
$d$ (diameter) = $[L^{1}]$
$\eta$ (coefficient of viscosity) = $[M^{1} L^{-1} T^{-1}]$
Substituting the dimensions into the formula:
$R_{e} = \frac{[M^{1} L^{-3} T^{0}] [M^{1} L^{1} T^{-1}] [L^{1}]}{[M^{1} L^{-1} T^{-1}]}$
$R_{e} = \frac{[M^{1} L^{-1} T^{-1}]}{[M^{1} L^{-1} T^{-1}]}$
$R_{e} = [M^{0} L^{0} T^{0}]$
Since all the powers of the fundamental dimensions are zero,the Reynolds number is dimensionless.

(B) The Reynolds number $(R_e)$ is a dimensionless quantity used to predict the flow regime of a fluid.
For flow through a pipe,the flow characteristics are generally defined as follows:
$1$. If $R_e < 2000$,the flow is laminar (streamline).
$2$. If $2000 < R_e < 4000$,the flow is in a transition state.
$3$. If $R_e > 4000$,the flow is turbulent.
However,in many simplified contexts,flows with $R_e > 1000$ to $2000$ are considered to be moving away from laminar flow and approaching turbulence. Given the options,$R_e > 1000$ indicates the onset of instability leading towards turbulent flow. In standard physics problems,$R_e > 2000$ is the strict threshold for turbulence,but $R_e > 1000$ is often associated with the transition to turbulent flow.

(TURBULENT) When the Reynolds number $(R_e)$ exceeds $2000$,the flow of the fluid is considered to be turbulent.
In turbulent flow,the fluid particles move in an irregular,chaotic,and unpredictable manner.
This type of flow is characterized by the formation of eddies,vortices,and fluctuations in velocity at any given point in the fluid.
Unlike laminar flow,where fluid layers slide smoothly over each other,turbulent flow involves significant mixing and energy dissipation due to internal friction.

(D) The Reynolds number $(R_e)$ is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
For steady (laminar) flow,$R_e < 2000$.
For transition flow,$2000 < R_e < 3000$.
For turbulent flow,$R_e > 3000$.
Unsteady flow refers to a flow where the velocity at a given point changes with time. The Reynolds number is specifically defined for steady-state flow conditions to characterize the regime of the flow. Therefore,it is not applicable or defined for unsteady flow.

(N/A) Critical velocity is the maximum velocity up to which the flow of a fluid remains streamlined or laminar. When the velocity of the fluid exceeds this value,the flow becomes turbulent.
The formula for critical velocity $(v_c)$ is given by:
$v_c = \frac{R_e \cdot \eta}{\rho \cdot D}$
Where:
$R_e$ is the Reynolds number (a dimensionless quantity),
$\eta$ is the coefficient of viscosity of the fluid,
$\rho$ is the density of the fluid,
$D$ is the diameter of the pipe.

(C) Turbulence is characterized by chaotic,stochastic property changes in fluid flow. The primary limitations are:
$1$. It is highly non-linear and sensitive to initial conditions,making long-term prediction impossible.
$2$. There is no single,universal analytical theory that describes all turbulent flows.
$3$. It involves a wide range of spatial and temporal scales,requiring massive computational power for numerical simulation $(DNS)$.
$4$. It significantly increases energy dissipation due to internal friction.

(N/A) Reynolds number is a dimensionless quantity used in fluid mechanics to predict flow patterns in different fluid flow situations.
It is defined as the ratio of inertial forces to viscous forces within a fluid which is subjected to relative internal movement due to different fluid velocities.
It depends on the density of the fluid $(\rho)$,the velocity of the fluid $(v)$,the characteristic length or diameter of the tube $(D)$,and the coefficient of viscosity $(\eta)$.
The formula for Reynolds number $(N_R)$ is:
$N_R = \frac{\rho v D}{\eta}$
Reynolds number is used to determine whether the flow of a liquid is laminar (streamline) or turbulent.

(A) The Reynolds number $(Re)$ is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
For streamline or laminar flow, the value of the Reynolds number is generally considered to be less than $2000$.
Flow is considered turbulent when $Re > 3000$, and it is in a transition state when $2000 < Re < 3000$.

(B) The Reynolds number $(Re)$ is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
For laminar flow, $Re < 1000$.
For unsteady or transitional flow, the Reynolds number typically lies between $1000$ and $2000$.
For turbulent flow, $Re > 2000$.
Therefore, the value of the Reynolds number for unsteady flow is between $1000$ and $2000$.

(C) The Reynolds number $(Re)$ is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
For flow in a pipe, the flow is considered laminar when $Re < 2000$.
The flow is considered transitional when $2000 < Re < 4000$.
The flow is considered turbulent when $Re > 4000$.
However, in many simplified contexts, flow is often categorized as turbulent when $Re > 2000$.

(B) In fluid dynamics,inertial force is the force associated with the momentum of the fluid particles in motion.
It is defined as the product of the mass of the fluid element and its acceleration.
Mathematically,it is represented as $F_i = m \cdot a$,where $m$ is the mass of the fluid and $a$ is the acceleration.
In the context of Reynolds number,it represents the resistance to change in the state of motion of the fluid.

(A) The Reynolds number $(Re)$ is a dimensionless quantity used in fluid mechanics to predict flow patterns in different fluid flow situations.
It is defined as the ratio of inertial forces to viscous forces within a fluid which is subjected to relative internal movement due to different fluid velocities.
Mathematically, it is expressed as:
$Re = \frac{\text{Inertial force}}{\text{Viscous force}}$
Therefore, the correct option is $A$.

(A) The Reynolds number $(R_e)$ is a dimensionless quantity used to predict the flow regime of a fluid.
$1$. For $R_e < 1000$,the flow is streamline or laminar.
$2$. For $1000 < R_e < 2000$,the flow is unsteady (transition state).
$3$. For $R_e > 2000$,the flow is turbulent.
Therefore,the correct matching is $(a-ii)$ and $(b-iii)$.

(D) The nature of flow is determined by the Reynolds number $(R_e)$: $R_e = \frac{\rho v D}{\eta}$.
Given: $\rho = 10^3\, kg/m^3$,$\eta = 10^{-3}\, Pa\cdot s$,$r = 0.5\, cm = 0.005\, m$,$D = 2r = 0.01\, m$.
Flow rate $Q = A \cdot v = \pi r^2 v$,so $v = \frac{Q}{\pi r^2}$.
Substituting $v$ in $R_e$: $R_e = \frac{\rho Q D}{\eta \pi r^2} = \frac{\rho Q (2r)}{\eta \pi r^2} = \frac{2 \rho Q}{\eta \pi r}$.
For $Q_1 = 0.18\, L/min = \frac{0.18 \times 10^{-3}}{60}\, m^3/s = 3 \times 10^{-6}\, m^3/s$:
$R_{e1} = \frac{2 \times 10^3 \times 3 \times 10^{-6}}{10^{-3} \times \pi \times 0.005} \approx 382.16$.
Since $R_{e1} < 1000$,the flow is steady.
For $Q_2 = 0.48\, L/min = \frac{0.48 \times 10^{-3}}{60}\, m^3/s = 8 \times 10^{-6}\, m^3/s$:
$R_{e2} = \frac{2 \times 10^3 \times 8 \times 10^{-6}}{10^{-3} \times \pi \times 0.005} \approx 1018.59$.
Since $1000 < R_{e2} < 2000$,the flow becomes unsteady.
Thus,the flow changes from steady to unsteady.