Answer
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Hint: In fluid mechanics, Venturimeter is a mechanical device which measures how much of fluid is flowing through a given time or we can say the rate of flow of a fluid through a pipe is measured by venturi meter and basically it measures in units of $litre{\min ^{ - 1}}$.
Complete step by step answer:
Let us suppose the area of first radii section which is $20\,cm$ is denoted by ${A_1} = \pi {(20)^2}\,c{m^2}$ and let the area of second radii section is denoted by ${A_2} = \pi {(10)^2}\,c{m^2}$
So now we have,
${A_1} = \pi {(20)^2}\,c{m^2}$
$\Rightarrow {A_2} = \pi {(10)^2}\,c{m^2}$
Acceleration due to gravity is $g = 980\,cm{\sec ^{ - 2}}$
And it’s given that the height up to which the water rises is $h = 10cm$
Volume flowing per second through the venturi meter is given by
$V = {A_1}{A_2}\sqrt {\dfrac{{2gh}}{{{A_1}^2 - {A_2}^2}}} $
In order to convert this flow of water in second to minutes and the cubic centimetres to litre,
Just multiply by $60$ we get,
$V = {A_1}{A_2}\sqrt {\dfrac{{2gh}}{{{A_1}^2 - {A_2}^2}}} \times 60\,litre\,{\min ^{ - 1}}$
Putting the values of all known parameters we will get:
$V = {(\dfrac{{22}}{7})^2}{(20)^2}{(10)^2}\sqrt {\dfrac{{2 \times 980 \times 10}}{{300(3.14)}}} \times 60litre{\min ^{ - 1}}$
$\Rightarrow V = {(\dfrac{{22}}{7})^2}2400000\sqrt {\dfrac{{19600}}{{942}}} litre{\min ^{ - 1}}$
$\therefore V = 2726.58\,litre\,{\min ^{ - 1}}$
Hence, the rate of flow of water measured by venturi meter in litres per minute is $V = 2726.58\,litre{\min ^{ - 1}}$.
Note: It should be remembered that, the basic units of conversions like: acceleration due to gravity is $g = 9.8m{\sec ^{ - 2}}$ into $g = 980cm{\sec ^{ - 2}}$ and litre is the most standard unit of volume which is used while measure large amount of fluids and its relation with cubic centimetre is given as $1litre = 1000c{m^3}$.
Complete step by step answer:
Let us suppose the area of first radii section which is $20\,cm$ is denoted by ${A_1} = \pi {(20)^2}\,c{m^2}$ and let the area of second radii section is denoted by ${A_2} = \pi {(10)^2}\,c{m^2}$
So now we have,
${A_1} = \pi {(20)^2}\,c{m^2}$
$\Rightarrow {A_2} = \pi {(10)^2}\,c{m^2}$
Acceleration due to gravity is $g = 980\,cm{\sec ^{ - 2}}$
And it’s given that the height up to which the water rises is $h = 10cm$
Volume flowing per second through the venturi meter is given by
$V = {A_1}{A_2}\sqrt {\dfrac{{2gh}}{{{A_1}^2 - {A_2}^2}}} $
In order to convert this flow of water in second to minutes and the cubic centimetres to litre,
Just multiply by $60$ we get,
$V = {A_1}{A_2}\sqrt {\dfrac{{2gh}}{{{A_1}^2 - {A_2}^2}}} \times 60\,litre\,{\min ^{ - 1}}$
Putting the values of all known parameters we will get:
$V = {(\dfrac{{22}}{7})^2}{(20)^2}{(10)^2}\sqrt {\dfrac{{2 \times 980 \times 10}}{{300(3.14)}}} \times 60litre{\min ^{ - 1}}$
$\Rightarrow V = {(\dfrac{{22}}{7})^2}2400000\sqrt {\dfrac{{19600}}{{942}}} litre{\min ^{ - 1}}$
$\therefore V = 2726.58\,litre\,{\min ^{ - 1}}$
Hence, the rate of flow of water measured by venturi meter in litres per minute is $V = 2726.58\,litre{\min ^{ - 1}}$.
Note: It should be remembered that, the basic units of conversions like: acceleration due to gravity is $g = 9.8m{\sec ^{ - 2}}$ into $g = 980cm{\sec ^{ - 2}}$ and litre is the most standard unit of volume which is used while measure large amount of fluids and its relation with cubic centimetre is given as $1litre = 1000c{m^3}$.
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