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Chapter 10: Problem 37

A cylinder containing water stands on a table of height \(H\). A small hole is punched in the side of cylinder at its base. The stream of water strikes the ground at a horizontal distance \(R\) from the table. Then the depth of water in the cylinder is (a) \(\vec{H}\) (b) \(R\) (c) \(\sqrt{R H}\) (d) \(R^{2} / 4 H\)

Short Answer

Expert verified
(d) \(\frac{R^{2}}{4H}\)

Step by step solution

01

Understanding the Problem

We have a cylinder with water and a hole at its base, causing water to stream out and hit the ground at a distance R. We need to determine the depth of water that leads to this scenario. We are given four possible options for the depth.
02

Using Torricelli's Law

Torricelli's Law relates the speed of efflux of a fluid through a hole to the height of the fluid above the hole. The speed of water coming out of the hole is given by \(v = \sqrt{2gh}\), where \(h\) is the depth of water above the hole and \(g\) is the acceleration due to gravity.
03

Calculating Time of Flight

The water takes time \(t\) to fall to the ground. The vertical distance to fall is the height of the table, \(H\). The time of flight is given by \(t = \sqrt{\frac{2H}{g}}\), using the equation for free fall under gravity.
04

Calculating Horizontal Distance

The horizontal distance \(R\) can be expressed as \(R = v \cdot t\). Substituting the expressions from Torricelli's Law and time of flight, we have \(R = \sqrt{2gh} \cdot \sqrt{\frac{2H}{g}}\).
05

Solving for Water Depth

Simplifying \(R = \sqrt{2gh} \cdot \sqrt{\frac{2H}{g}}\) gives \(R = \sqrt{4hH}\). This simplifies to \(R^2 = 4hH\), leading to \(h = \frac{R^2}{4H}\).
06

Selecting the Correct Option

The depth of water in the cylinder is \(\frac{R^2}{4H}\), which matches option (d).

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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 pivotal in fluid mechanics, especially when analyzing how liquids behave under the influence of gravity. It provides a way to determine the velocity of a fluid flowing out through an opening. This velocity is determined by the height of the fluid column above the opening. The fundamental formula derived from Torricelli's Law is:
\[ v = \sqrt{2gh} \]
where:
  • \(v\) is the velocity of the fluid outflow.
  • \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\) on Earth.
  • \(h\) is the height of the fluid above the hole.
This law assumes that the fluid is incompressible and that the flow is steady. It also presumes there is no energy loss in the flow, meaning factors like viscosity are ignored. Torricelli's Law is derived from Bernoulli鈥檚 principle, which itself is a conservation of energy statement for flowing fluids. It beautifully demonstrates how gravitational potential energy is converted to kinetic energy as the fluid exits the hole.
Stream Flow
Stream flow in this context refers to how water exits the cylinder through a small hole at its base and travels through the air until it hits the ground. When the water is allowed to exit, it undergoes two components of motion:
  • **Horizontal flow** 鈥� dictated by the velocity from Torricelli's Law.
  • **Vertical free-fall** 鈥� as gravity pulls the water downwards.
In analyzing the stream flow, we use the time of flight (\(t\)), which tells us how long it takes for the water to reach the ground from the height of the table \(H\). This can be calculated using the formula:
\[ t = \sqrt{\frac{2H}{g}} \]
This formula results from solving the equation of motion under constant acceleration due to gravity. For the horizontal distance \(R\) traveled by the stream, we use:
\[ R = v \cdot t \]
By substituting \(v\) from Torricelli鈥檚, we discovered that this stream鈥檚 reach is proportional to both the speed at which it exits the cylinder and the time it takes to fall.
Gravity
Gravity plays a vital role in fluid mechanics, especially in the context of this exercise. It acts as the continuous force pulling the water downward after it exits the hole. Its influence is accounted for when determining:
  • The velocity of the fluid using Torricelli鈥檚 Law.
  • The time of flight during which the fluid falls to the ground.
  • The trajectory path that the fluid liquid follows.
Gravity is approximately \(9.8 \text{ m/s}^2\), meaning it accelerates all objects towards the Earth's surface at this rate. When fluid is in free fall, it behaves according to the equations of motion. As it leaves the cylinder, the stream鈥檚 vertical motion is a result of gravitational pull, while the horizontal motion results from the initial velocity imparted by leaving the hole.
In the exercise, after you calculate the height \(h\) using the relationship \(R^2 = 4hH\), it's clear that gravity is the central player converting potential energy in the fluid to kinetic energy in the stream.

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Most popular questions from this chapter

A large tank filled with water to a height of \(h\) is said to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall down from \(h\) to \(h / 2\) and from \(h / 2\) to zero is (a) \(\sqrt{2}\) (b) \(1 / \sqrt{2}\) (c) \(\sqrt{2}-1\) (d) \(\frac{1}{\sqrt{2}-1}\)

If the terminal speed of a sphere of gold (density \(=19.5\) \(\mathrm{kg} / \mathrm{m}^{3}\) ) is \(0.2 \mathrm{~m} / \mathrm{s}\) in a viscous liquid \(\left(\right.\) density \(\left.=1.5 \mathrm{~kg} / \mathrm{m}^{3}\right)\) find the terminal speed of a sphere of silver (density = \(10.5 \mathrm{~kg} / \mathrm{m}^{3}\) ) of the same size in the same liquid (a) \(0.2 \mathrm{~m} / \mathrm{s}\) (b) \(0.4 \mathrm{~m} / \mathrm{s}\) (c) \(0.133 \mathrm{~m} / \mathrm{s}\) (d) \(0.1 \mathrm{~m} / \mathrm{s}\)

A spherical ball of radius \(r\) and relative density \(0.5\) is floating in equilibrium in water with half of it immersed in water. The work done in pushing the ball down so that whole of it is just immersed in water is (where \(\rho\) is the density of water) (a) \(\frac{5}{12} \pi r^{4} \rho g\) (b) \(0.5 \rho r g\) (c) \(\frac{4}{3} \pi r^{3} \rho g\) (d) \(\frac{2}{3} \pi r^{4} \rho g\)

A tank is filled with water upto a height \(H\). Water is allowed to come out of a hole \(P\) in one of the walls at a depth \(D\) below the surface of water. Express the horizontal distance \(x\) in terms of \(H\) and \(D\) (a) \(x=\sqrt{D(H-D)}\) (b) \(x=\sqrt{\frac{D(H-D)}{2}}\) (c) \(x=2 \sqrt{D(H-D)}\) (d) \(x=4 \sqrt{D(H-D)}\)

A body floats in water with one-third of its volume above the surface of water. If it is placed in oil, it floats with half of its volume above the surface of the oil. The specific gravity of the oil is (a) \(\frac{5}{3}\) (b) \(\frac{4}{3}\) (c) \(\frac{3}{2}\) (d) 1
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