
In a plant, sucrose solution of coefficient of viscosity $0.0015\;{\rm{N}}{{\rm{m}}^{ - 2}}$ is driven at a velocity of ${10^{ - 3}}\;{\rm{m}}{{\rm{s}}^{ - 1}}$ through xylem vessels of radius ${\rm{2}}\;{\rm{\mu m}}$ and length ${\rm{5}}\;{\rm{\mu m}}$. The hydrostatic pressure difference across the length of xylem vessels in ${\rm{N}}{{\rm{m}}^{ - 2}}$ is:
(A) $5$
(B) $8$
(C) $10$
(D) $15$
Answer
574.5k+ views
Hint: In every question, we should first identify the given quantities. Then, we should try to obtain an equation where the given quantities can be substituted. In this question, Poiseuille’s formula for the volume flow rate can be used to find the solution.
Complete step by step answer:
Given:
The coefficient of viscosity, $\eta = 0.0015\;{\rm{N}}{{\rm{m}}^{ - 2}}$
The velocity with which the sucrose solution moves through the xylem vessels, $v = {10^{ - 3}}\;{\rm{m}}{{\rm{s}}^{ - 1}}$
The radius of the xylem vessel,
\[
r = 2\;{\rm{\mu m}}\\
\Rightarrow r = 2 \times {10^{ - 6}}\;{\rm{m}}
\]
The length of the xylem vessel,
\[
l = {\rm{5}}\;{\rm{\mu m}}\\
\Rightarrow l = {\rm{5}} \times {\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{m}}
\]
Now, the volume flow rate of the solution flowing through the xylem vessel is given by
$Q = Av$
Here $Q$ is the volume flow rate of the solution, $A$ is the area of the cross section of the vessel and $v$ is the velocity of flow of the solution.
Since the cross section of the xylem vessel is circular in shape, the area of cross section is given by,
$A = \pi {r^2}$
We can insert this equation in $Q = Av$. Thus, we get
$Q = \pi {r^2}v$
Since the solution flows in layers and in a smooth way, we can say that the flow is laminar.
So, now Poiseuille’s formula for the volume flow rate can be written as
$Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}$
Here we define $P$ as the hydrostatic pressure difference, $r$ as the radius of the vessel, $\eta $ as the coefficient of viscosity and $l$ as the length of the vessel.
So, now we have got two equations for the volume flow rate. We can equate both the equations to obtain an equation to find the hydrostatic pressure difference. Equating equations $Q = \pi {r^2}v$ and $Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}$, we get
$
\dfrac{{\pi {{\Pr }^4}}}{{8\eta l}} = \pi {r^2}v\\
\Rightarrow \dfrac{{{{\Pr }^2}}}{{8\eta l}} = v\\
\Rightarrow P = \dfrac{{8\eta lv}}{{{r^2}}}
$
Now, we just have to substitute the values of $\eta $, $l$, $v$ and $r$ in the obtained equation to find the pressure difference. Substituting the values, we get
$
P = \dfrac{{8 \times 0.0015\;{\rm{N}}{{\rm{m}}^{ - 2}} \times {\rm{5}} \times {\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{m}} \times {{10}^{ - 3}}\;{\rm{m}}{{\rm{s}}^{ - 1}}}}{{{{\left( {2 \times {{10}^{ - 6}}\;{\rm{m}}} \right)}^2}}}\\
\therefore P = 15\;{\rm{N}}{{\rm{m}}^{{\rm{ - 2}}}}
$
Hence, the value of the hydrostatic pressure difference is obtained as $15\;{\rm{N}}{{\rm{m}}^{{\rm{ - 2}}}}$. Therefore, option (D) is correct.
Note: We can use Poiseuille’s formula to find the volume flow rate of fluids only if it is laminar flow. For the turbulent flow of fluids, Poiseuille’s formula is not applicable.
Complete step by step answer:
Given:
The coefficient of viscosity, $\eta = 0.0015\;{\rm{N}}{{\rm{m}}^{ - 2}}$
The velocity with which the sucrose solution moves through the xylem vessels, $v = {10^{ - 3}}\;{\rm{m}}{{\rm{s}}^{ - 1}}$
The radius of the xylem vessel,
\[
r = 2\;{\rm{\mu m}}\\
\Rightarrow r = 2 \times {10^{ - 6}}\;{\rm{m}}
\]
The length of the xylem vessel,
\[
l = {\rm{5}}\;{\rm{\mu m}}\\
\Rightarrow l = {\rm{5}} \times {\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{m}}
\]
Now, the volume flow rate of the solution flowing through the xylem vessel is given by
$Q = Av$
Here $Q$ is the volume flow rate of the solution, $A$ is the area of the cross section of the vessel and $v$ is the velocity of flow of the solution.
Since the cross section of the xylem vessel is circular in shape, the area of cross section is given by,
$A = \pi {r^2}$
We can insert this equation in $Q = Av$. Thus, we get
$Q = \pi {r^2}v$
Since the solution flows in layers and in a smooth way, we can say that the flow is laminar.
So, now Poiseuille’s formula for the volume flow rate can be written as
$Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}$
Here we define $P$ as the hydrostatic pressure difference, $r$ as the radius of the vessel, $\eta $ as the coefficient of viscosity and $l$ as the length of the vessel.
So, now we have got two equations for the volume flow rate. We can equate both the equations to obtain an equation to find the hydrostatic pressure difference. Equating equations $Q = \pi {r^2}v$ and $Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}$, we get
$
\dfrac{{\pi {{\Pr }^4}}}{{8\eta l}} = \pi {r^2}v\\
\Rightarrow \dfrac{{{{\Pr }^2}}}{{8\eta l}} = v\\
\Rightarrow P = \dfrac{{8\eta lv}}{{{r^2}}}
$
Now, we just have to substitute the values of $\eta $, $l$, $v$ and $r$ in the obtained equation to find the pressure difference. Substituting the values, we get
$
P = \dfrac{{8 \times 0.0015\;{\rm{N}}{{\rm{m}}^{ - 2}} \times {\rm{5}} \times {\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{m}} \times {{10}^{ - 3}}\;{\rm{m}}{{\rm{s}}^{ - 1}}}}{{{{\left( {2 \times {{10}^{ - 6}}\;{\rm{m}}} \right)}^2}}}\\
\therefore P = 15\;{\rm{N}}{{\rm{m}}^{{\rm{ - 2}}}}
$
Hence, the value of the hydrostatic pressure difference is obtained as $15\;{\rm{N}}{{\rm{m}}^{{\rm{ - 2}}}}$. Therefore, option (D) is correct.
Note: We can use Poiseuille’s formula to find the volume flow rate of fluids only if it is laminar flow. For the turbulent flow of fluids, Poiseuille’s formula is not applicable.
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