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I will derive the Reynolds Number denoting all the variables along the way

$A \rightarrow$ Area of cross section of tube

$v \rightarrow$ Velocity of fluid flow in tube

$\rho \rightarrow$ Density of fluid in tube

$r \rightarrow$ Radius of tube

$\eta \rightarrow$ Coefficient of viscosity

Mass of fluid flowing per second = $\mu$ = $\rho Av$

Inertial force = Force due to change in momentum = $\Delta l$= $\mu \cdot v$ = $\rho Av^2$

Viscous force = $F$ = $\eta \frac{Av}{r}$

Reynolds Number = $\frac{\Delta l}{F} = \frac{\rho A v^2}{\eta \frac{Av}{r}} = \frac{\rho A v^2 r}{\eta A v}$

Now this is of the form that was given in NCERT. which means that $A$ was cross sectional area after all.

I will derive the Reynolds Number denoting all the variables along the way

$A \rightarrow$ Area of cross section of tube

$v \rightarrow$ Velocity of fluid flow in tube

$\rho \rightarrow$ Density of fluid in tube

$r \rightarrow$ Radius of tube

$\eta \rightarrow$ Coefficient of viscosity

Mass of fluid flowing per second = $\mu$ = $\rho Av$

Inertial force = Force due to change in momentum = $\Delta l$= $\mu \cdot v$ = $\rho Av^2$

Viscous force = $F$ = $\eta \frac{Av}{r}$

Reynolds Number = $\frac{\Delta l}{F} = \frac{\rho A v^2}{\eta \frac{Av}{r}} = \frac{\rho A v^2 r}{\eta A v}$

Now this is of the form that was given in NCERT. which means that $A$ was cross sectional after all.

I will derive the Reynolds Number denoting all the variables along the way

$A \rightarrow$ Area of cross section of tube

$v \rightarrow$ Velocity of fluid flow in tube

$\rho \rightarrow$ Density of fluid in tube

$r \rightarrow$ Radius of tube

$\eta \rightarrow$ Coefficient of viscosity

Mass of fluid flowing per second = $\mu$ = $\rho Av$

Inertial force = Force due to change in momentum = $\Delta l$= $\mu \cdot v$ = $\rho Av^2$

Viscous force = $F$ = $\eta \frac{Av}{r}$

Reynolds Number = $\frac{\Delta l}{F} = \frac{\rho A v^2}{\eta \frac{Av}{r}} = \frac{\rho A v^2 r}{\eta A v}$

Now this is of the form that was given in NCERT. which means that $A$ was cross sectional area after all.

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I will derive the Reynolds Number denoting all the variables along the way

$A \rightarrow$ Area of cross section of tube

$v \rightarrow$ Velocity of fluid flow in tube

$\rho \rightarrow$ Density of fluid in tube

$r \rightarrow$ Radius of tube

$\eta \rightarrow$ Coefficient of viscosity

Mass of fluid flowing per second = $\mu$ = $\rho Av$

Inertial force = Force due to change in momentum = $\Delta l$= $\mu \cdot v$ = $\rho Av^2$

Viscous force = $F$ = $\eta \frac{Av}{r}$

Reynolds Number = $\frac{\Delta l}{F} = \frac{\rho A v^2}{\eta \frac{Av}{r}} = \frac{\rho A v^2 r}{\eta A v}$

Now this is of the form that was given in NCERT. which means that $A$ was cross sectional after all.