Answer
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Hint: The ratio of the inertial forces to the viscosity force that is subjected to the internal movement which is related to the different fluid velocities is known as the Reynolds number. It is a dimensionless number that is used to determine the types of flow patterns which are laminar or turbulent.
Formula used:
To find the Reynolds number,
${R_e} = \dfrac{{\rho vL}}{\mu }$
Where,
$\rho $ is the density,
$v$ is the speed of the flow,
$L$ is the linear dimension characteristic,
$\mu $is the dynamic viscosity
Complete step by step answer:
The values are given in the question. The liquid of coefficient of the viscosity $\eta = 1$ poise, pipe of the radius $3cm$, and the rate of volume flow are $1000l/\min $.
The rate of the flow is $1000l/\min $
The value of $\pi {r^2}$is $\dfrac{1}{{60}}{m^3}/s$
Substitute all the values in the Reynolds formula. We have,
${R_e} = \dfrac{{\rho vL}}{\mu }$
Where,
$\rho $ is the density,
$v$ is the speed of the flow,
$L$ is the linear dimension characteristic,
$\mu $ is the dynamic viscosity
$ \Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{1}{{60\pi {r^2}}} \times 2r$
$ \Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{{2r}}{{60\pi {r^2}}}$
$ \Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{2}{{60\pi r}}$
$ \Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\pi \times r}}$
The value of the radius is $3cm$ converting the centimeter into the meter we get,
$ \Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\pi \times 3 \times {{10}^{ - 2}}}}$
Substituting the value of $\pi$,
$ \Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\left( {3.16} \right) \times 3 \times {{10}^{ - 2}}}}$
\[ \Rightarrow {R_e} = \dfrac{{2000}}{{568.8 \times {{10}^{ - 2}}}}\]
\[ \Rightarrow {R_e} = 3563\]
Therefore the Reynolds number is 3563.
Hence option \[\left( A \right)\] is the correct answer
Note: If the Reynolds number has a high value the pipe has a turbulent flow. If the Reynolds number has a low value the pipe has a laminar flow. Numerically these values are acceptable though in general the laminar and turbulent flow can be classified according to the range.
Formula used:
To find the Reynolds number,
${R_e} = \dfrac{{\rho vL}}{\mu }$
Where,
$\rho $ is the density,
$v$ is the speed of the flow,
$L$ is the linear dimension characteristic,
$\mu $is the dynamic viscosity
Complete step by step answer:
The values are given in the question. The liquid of coefficient of the viscosity $\eta = 1$ poise, pipe of the radius $3cm$, and the rate of volume flow are $1000l/\min $.
The rate of the flow is $1000l/\min $
The value of $\pi {r^2}$is $\dfrac{1}{{60}}{m^3}/s$
Substitute all the values in the Reynolds formula. We have,
${R_e} = \dfrac{{\rho vL}}{\mu }$
Where,
$\rho $ is the density,
$v$ is the speed of the flow,
$L$ is the linear dimension characteristic,
$\mu $ is the dynamic viscosity
$ \Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{1}{{60\pi {r^2}}} \times 2r$
$ \Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{{2r}}{{60\pi {r^2}}}$
$ \Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{2}{{60\pi r}}$
$ \Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\pi \times r}}$
The value of the radius is $3cm$ converting the centimeter into the meter we get,
$ \Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\pi \times 3 \times {{10}^{ - 2}}}}$
Substituting the value of $\pi$,
$ \Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\left( {3.16} \right) \times 3 \times {{10}^{ - 2}}}}$
\[ \Rightarrow {R_e} = \dfrac{{2000}}{{568.8 \times {{10}^{ - 2}}}}\]
\[ \Rightarrow {R_e} = 3563\]
Therefore the Reynolds number is 3563.
Hence option \[\left( A \right)\] is the correct answer
Note: If the Reynolds number has a high value the pipe has a turbulent flow. If the Reynolds number has a low value the pipe has a laminar flow. Numerically these values are acceptable though in general the laminar and turbulent flow can be classified according to the range.
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