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A cylindrical bucket, open at the top, is 25.0 \(\mathrm{cm}\) high and 10.0 \(\mathrm{cm}\) in diameter. A circular hole with a cross-sectional area 1.50 \(\mathrm{cm}^{2}\) is cutin the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of \(2.40 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}\) . How high will the water in the bucket rise?

Short Answer

Expert verified
The water will rise to 13.06 cm in the bucket.

Step by step solution

01

Convert Units

Convert all measurements to consistent units. The diameter of the bucket is given in centimeters so convert it to meters: 10 cm = 0.1 m. The height of the bucket is 25 cm which is 0.25 m. The cross-sectional area of the hole is given as 1.50 cm², which is 0.00015 m².
02

Calculate the Area of the Top Opening of the Bucket

The top opening is circular, so use the formula for the area of a circle: \[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.1}{2}\right)^2 = 0.00785 \ m^2 \].
03

Relation Between Water Inflow and Outflow at Equilibrium

The inflow rate is given as \(2.40 \times 10^{-4} \ m^3/s \). The rate at which water exits through the hole depends on the area of the hole and the velocity of the water, governed by Torricelli's Law: \(v = \sqrt{2gh}\), which states that the velocity depends on the height \(h\) of the water in the bucket.
04

Use Torricelli's Law to Eliminate Velocity

Express the outflow rate using the area of the hole \(A_{hole}\) and the velocity \(v\) as follows: \( \text{Outflow rate} = A_{hole} \cdot v = A_{hole} \cdot \sqrt{2gh} \). Substitute \( v = \sqrt{2gh} \) into the inflow-rate equalization: \[2.40 \times 10^{-4} = 0.00015 \cdot \sqrt{2gh} \].
05

Solve for Height h

Isolate \( h \) in the equation from the previous step: \[ \sqrt{2gh} = \frac{2.40 \times 10^{-4}}{0.00015} \].Now, square both sides to remove the square root:\[ 2gh = \left(\frac{2.40 \times 10^{-4}}{0.00015}\right)^2 \]\[ h = \frac{1}{2g} \left(\frac{2.40 \times 10^{-4}}{0.00015}\right)^2 \]. Plug in values, noting \( g = 9.81 \ m/s^2 \): \[ h \approx \frac{1}{2 \times 9.81} \left(1.6 \right)^2 \approx 0.1306 \].
06

Check Against Bucket Height

The calculated height of the water, \( h \approx 0.1306 \ m \) or 13.06 cm, is less than the height of the bucket's side (25 cm), so overflow does not occur. This means the bucket can maintain this water height at equilibrium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torricelli's Law
Torricelli's Law is a fundamental principle in fluid dynamics that enables us to determine the speed of fluid flowing out of an opening due to gravity. Named after the Italian physicist Evangelista Torricelli, the law simplifies the analysis of fluid flow by relating the velocity of the fluid to the height of the fluid column above the opening. According to Torricelli's Law, the velocity (\(v\)) of the exiting fluid can be expressed as \(v = \sqrt{2gh}\), where \(g\) is the acceleration due to gravity (approximately 9.81 \(m/s^2\)) and \(h\) is the height of the fluid column in meters.

This law is particularly useful for problems involving fluid flowing out of a hole or opening, where the height of the fluid is a significant factor that influences flow speed. By using Torricelli’s Law, we can effectively connect the dynamics of fluid pressure and gravitational forces to real-world flow conditions. This is essential for understanding how changes in height can affect fluid flow rates in systems like tanks, buckets, or reservoirs.
Flow Rate
The flow rate of a fluid refers to the volume of fluid that passes a point per unit time, typically expressed as \(m^3/s\) or liters per second (L/s). It is an essential quantity in fluid dynamics, representing how fast or slow a fluid is moving within a system. In a practical context, like our bucket problem, understanding flow rate helps in predicting the behavior of the fluid within the container.

In this scenario, the inflow rate is provided as \(2.40 \times 10^{-4} \ m^3/s\). This indicates how quickly the water fills the bucket. Meanwhile, the outflow rate is determined by the area of the hole and Torricelli’s calculated velocity. By ensuring that inflow and outflow rates balance, we can achieve a steady-state condition where the height of the water remains constant.

The relationship between inflow and outflow is crucial for figuring out equilibrium conditions in systems where fluids are being added and removed simultaneously. It's instrumental in designing systems or solving problems involving continuous fluid processes.
Unit Conversion
Unit conversion is a crucial step in solving any physics problem because it allows us to compare and calculate accurately by ensuring every quantity is in the same set of units. For this problem, all measurements are originally given in centimeters and centimeters squared, which are not standard units in physics, especially when compared with the inflow rate given in cubic meters per second.

To maintain consistency, the diameter and height of the bucket are converted from centimeters to meters:
  • The diameter: 10 cm = 0.1 m
  • The height: 25 cm = 0.25 m
Similarly, the area of the hole, initially given as 1.50 cm\(^2\), is converted into square meters: 0.00015 m\(^2\).

Converting units into the International System of Units (SI) allows for straightforward application of physical laws and formulas. This results in easier calculations and reduces the chance of mistakes that can occur with mixed units. Correct unit conversion is a fundamental skill in physics that ensures consistency and precision in calculations.
Water Pressure
Water pressure is another key element in fluid dynamics and it influences how fluids behave under various conditions. In the context of our bucket problem, water pressure plays a critical role as it is related to the height of the water column (\(h\)). Water pressure at a given depth emerges from the weight of the water above it and can be calculated using the formula \(P = \rho gh\), where \(\rho\) is the density of the fluid (typically water, which is 1000 kg/m^3).

This pressure then exerts force on the fluid at the hole, thereby influencing the velocity at which water exits according to Torricelli's Law. In essence, the taller the water column, the greater the pressure at the bottom, and the faster the exit velocity of the water from the bucket. Understanding the impact of water pressure is crucial for tackling problems that involve fluid discharge, and it's invariably tied to the flow rate and fluid height within the system. This helps predict the system's behavior under various operational conditions.

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Most popular questions from this chapter

A tall cylinder with a cross-sectional area 12.0 \(\mathrm{cm}^{2}\) is partially filled with mercury; the surface of the mercury is 5.00 \(\mathrm{cm}\) above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder?

Miles per Kilogram. The density of gasoline is 737 \(\mathrm{kg} / \mathrm{m}^{3}\) . If your new hybrid car gets 45.0 miles per gallon of gasoline, what is its mileage in miles per kilogram of gasoline? (See Appendix E.)

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