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A pipe having an internal diameter D is connected to a pipe of the same size. Water flows into the second pipe through n holes, each of diameter d. If the water in the first pipe has speed v, the speed of water leaving the second pipe is:
A.$\dfrac{{{D^2}v}}{{n{d^2}}}$
B.$\dfrac{{n{D^2}v}}{{{d^2}}}$
C.$\dfrac{{n{d^2}v}}{{{D^2}}}$
D.$\dfrac{{{d^2}v}}{{n{D^2}}}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Volume of water flowing per second through the first pipe is same as the volume of water flowing per second from all the holes

Step by step answer: The pipe whose internal diameter is D is connected to another pipe of the same size. Water from the first pipe enters the second pipe through n holes each having diameter d. The velocity of water flowing in the first pipe is v and let the velocity of water leaving through each hole be vse , R be the radius of the first pipe and r be the radius of each hole.
Area of cross section of first pipe (A)=$\pi {R^2}$
$\pi {\left( {\dfrac{D}{2}} \right)^2} = \left[ {radius = \dfrac{{diameter}}{2}} \right]$
Area of cross section of each hole (a) = $\pi {r^2} = \pi {\left( {\dfrac{d}{2}} \right)^2}$
Volume of water flowing per second through First pipe = Av$\left[ {volume = area \times velocity} \right]$

Mass of water flowing per second through first pipe = $volume \times density$
 $Av \times \rho $ where ρ is the density of water

Volume of water flowing per second through each hole = $a\left( {vse} \right)$
Mass of water flowing per second through n number of holes = $n \times a \times vse \times \rho $

The equation $Av = a\left( {vse} \right)$is known as the equation of continuity.
As there are no sources or sinks in the pipe where water can be created or destroyed. So, the mass of water flowing per second is the same. Hence , $A \times v \times \rho = n \times a \times vse \times \rho $

${\left( {\dfrac{D}{2}} \right)^2}v = n{\left( {\dfrac{d}{2}} \right)^2}vse$
$\Rightarrow$ $\dfrac{{{D^2} \times v}}{4} = \left( {\dfrac{{n \times {d^2} \times v}}{4}} \right)vse$
$\Rightarrow$ $vse = \dfrac{{v \times {D^2} \times 4}}{{n \times {d^2} \times 4}}$
$\therefore$ $vse = \dfrac{{v \times {D^2}}}{{n \times {d^2}}}$ $ = \dfrac{{v{D^2}}}{{n{d^2}}}$
The velocity of water flowing through the second pipe is the same as the velocity of water through the hole.
Therefore, option A is correct.

Note: Since the mass of water flowing through the first pipe and n number of holes are the same. So, the mass of water flowing per second in both the cases is the same.