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This is the second of our many quizzes on Volume flow rate. Click on this to access more such competitive quizzes on Mechanics.

Volume flow rate Quiz 2

physics

The volume flow rate in terms of mass flow rate $\dot{m}$ is,

Mass flow rate $\dot{m}$ is the mass of the liquid that passes through per unit time and given by,$\,\dot{m}=Q\cdot \rho $. It is used to determine the volume flow rate when density $\rho $ is constant and $\dot{m}$ the mass flow rate is given. 

  $\text{Mass flow rate}\,\dot{m}\,\text{is the mass of the liquid that passes through per unit time,}\\ \text{and when given can be used to determine the volume flow rate as density is constant,}\\ Q=\dfrac{{\dot{m}}}{\rho }\\ \text{Where}\,\dot{m}\,\text{is the mass flow rate in kg/s and}\, \rho\,\text{is the density of the liquid in}\, kg/{{m}^{3}}$.

Petrol flows through a pipe of radius $2$ m at a velocity of $20$ m/s. What will be the velocity of petrol if the pipe narrows to a radius of $1$ m and is not compressed? 

Since petro is not compressed the amount of petro that flows in the pipe should be equal to the amount of petro that comes out of the pipe, use the equation of continuity given by ${{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}$. Plug in the known values and find the unknown velocity. 

  $\text{Petro is not compressed, so the amount of petro that flows in the pipe should be equal to}\\ \text{the amount of petro that comes out of the pipe.}\\ \text{This is called the continuity of flow and is given by the equation,}\\ {{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}….(i)\\ \text{According to the question,}\\ {{r}_{1}}=2m,{{r}_{2}}=1m\,\text{and}\,{{v}_{1}}=20\,m/s\\ \text{Putting the above values in equation (i), we get,}\\ \pi \times {{(2)}^{2}}\times 20=\pi \times {{(1)}^{2}}\times {{v}_{2}}\\ \therefore {{v}_{2}}=80\, m/s$ 

What is the volume of blood pumped in $42$ years if volume flow rate is $5$ L/min? 

 The volume of flow is given by $V=Qt$.By plugging in the values of time in seconds and volume flow rate in ${{m}^{3}}/s$ in the above equation we can calculate the volume. 

 $\text{Volume flow rate is defined by,}\\ Q=\dfrac{V}{t}\\ \Rightarrow V=Qt….(i)\\ \text{According to the question,}\\ Q=5\,L/min=\dfrac{5}{1000\times 60}{{m}^{3}} /s\,\\ \text{And}\,t=(42\times 60\times 60\times 24\times 365)\,seconds\\ \text{Putting the above values in equation (i), we get,}\\ \therefore V=1.1\times {{10}^{5}}\,{{m}^{3}}$

Methane flows from a pipe of diameter $10$ m with a speed of $13$ m/s to a network of identical smaller pipes of radius $1$ m with a speed of $5$ m/s.How many such pipes are in the network? 

Since methane is not compressed, the amount of methane that flows in the pipe should be equal to the amount of methane that comes out all the smaller pipes taken together, use the equation of continuity given by ${{n}_{1}}{{A}_{1}}{{v}_{1}}={{n}_{2}}{{A}_{2}}{{v}_{2}}$. Plug in the known values and find ${{n}_{2}}$. 

 $\text{Methane is not compressed inside the pipe,}\\ \text{so the amount of methane that flows in the pipe should be equal to}\\ \text{the amount of methane that comes out of all the smaller pipes.}\\ \text{Thus, the continuity equation is given by,}\\ {{n}_{1}}{{A}_{1}}{{v}_{1}}={{n}_{2}}{{A}_{2}}{{v}_{2}}….(i)\\ \text{According to the question,}\\ {{n}_{1}}=1,{{r}_{1}}=\dfrac{10}{2}=5m, {{r}_{2}}=1m,{{v}_{1}}=13\,m/s\\ \text{And}\,{{v}_{2}}=5\,m/s\\ \text{Putting the above values in equation (i), we get,}\\ 1\times \pi \times {{(5)}^{2}}\times 13={{n}_{2}}\times \pi \times {{(1)}^{2}}\times 5 \\ \Rightarrow {{n}_{2}}=65$

Water is flowing in a T-shaped pipe as shown in the figure below. What is $V$ if ${{V}_{1}}$ and ${{V}_{2}}$ are $4$ m/s and $6$ m/s, respectively. $(\text{Assuming water is not compressed})$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/64a593ea-8461-4931-b453-b06c1faa360a6086550719536466237.png"/>

Since water is not compressed the amount of water that flows the two pipes should be equal to the amount of water that comes out of the pipe, use the equation of continuity given by ${{A}_{1}}{{v}_{1}}+{{A}_{2}}{{v}_{2}}={{A}_{3}}v$. Plug in the known values and find the unknown velocity. 

 $\text{Water is not compressed,}\\ \text{so the amount of water that flows in through the two pipes should be equal to}\\ \text{the amount of water that comes out of the pipe.}\\ \text{Hence, here the continuity equation becomes,}\\ 2A{{V}_{1}}+2A{{V}_{2}}=AV    ….(i)\\ \text{According to the question,}\\ {{V}_{1}}=4\, m/s\,\text{and}\,{{V}_{2}}=6\,m/s\\ \text{Putting the above values in equation (i), we get,}\\ \therefore V=20\,m/s$ 