The cylindrical tube of a spare pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min-1, what is the speed of ejection of the liquid through the holes?

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To solve this problem, we need to consider the principle of conservation of volume flow rate. The volume flow rate of the liquid inside the tube must be equal to the sum of the volume flow rates through each of the fine holes.

The cross-sectional area of the cylindrical tube is 8.0 cm28.0cm2. Therefore, the area AA of one fine hole can be calculated using the formula for the area of a circle:

A=π×r2A=π×r2

Given that the diameter of each hole is 1.0 mm1.0mm, the radius rr is 0.5 mm0.5mm or 0.05 cm0.05cm. Hence, the area of one fine hole is:

A=π×(0.05)2 cm2A=π×(0.05)2cm2

Now, we can find the total area of all 40 fine holes:

Total area (Atotal)=40×ATotalarea(Atotal)=40×A

The volume flow rate QQ through each hole can be calculated using the formula:

Q=A×vQ=A×v

Where vv is the speed of ejection of the liquid through the holes. Since the volume flow rate through each hole is the same, we can write:

Qtotal=40×QQtotal=40×Q

Given that the liquid flow inside the tube is 1.5 m min−11.5m min−1, we first need to convert this into cm/min:

1.5 m min−1=150 cm min−11.5m min−1=150cm min−1

Now, we substitute the values into the equation:

Qtotal=150×8.0=1200 cm3min−1Qtotal=150×8.0=1200cm3min−1

Since Qtotal=40×QQtotal=40×Q, we can solve for QQ as:

Q=Qtotal40Q=40Qtotal

Q=120040=30 cm3min−1Q=401200=30cm3min−1

Now, we substitute the value of QQ into the equation Q=A×vQ=A×v to find the speed of ejection (vv):

30=(0.05)2×π×v30=(0.05)2×π×v

Solving for vv, we get:

v=30π×(0.05)2v=π×(0.05)230

v≈300.00785v≈0.0078530

v≈3821.66 cm min−1v≈3821.66cm min−1

So, the speed of ejection of the liquid through the holes is approximately 3821.66 cm min−13821.66cm min−1.

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