/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A liquid of density \(\rho\) is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 6

A liquid of density \(\rho\) is completely filled in a rectangular box. The box is accelerating horizontally with acceleration \(a\). What should be the gauge pressure at four points \(P, Q, R, S ?\) (a) \(P_{P}=0, P_{Q}=0, P_{R}=\rho g h, P_{S}=0\) (b) \(P_{P}=\rho g h, P_{o}=0, P_{k}=\rho g a, P_{s}=\rho a L\) (c) \(P_{P}=0, P_{Q}=\rho g h, P_{k}=\rho g h+\rho a L, P_{S}=\rho a L\) (d) \(P_{P}=\rho g h, P_{\varrho}=0, P_{R}=\rho g h-\rho a L, P_{s}=\rho g L\)

Short Answer

Expert verified
The correct option is (c).

Step by step solution

01

Understanding Gauge Pressure

Gauge pressure at any point in a fluid is the difference between the pressure at that point and atmospheric pressure. This problem involves a fluid in a box accelerating horizontally, so we have to consider both gravitational forces and the effect of the horizontal acceleration on the fluid.
02

Analyzing Point P

Point P is at the top of the box where the liquid level is open to the atmosphere. Therefore, since gauge pressure is measured relative to atmospheric pressure, the gauge pressure at point P is 0.
03

Analyzing Point Q

Point Q is deeper in the liquid directly under point P. At this point, the pressure would be due to the weight of the liquid above. Hence, the gauge pressure is calculated as \(P_{Q} = \rho g h\), where \(h\) is the depth of the liquid above point Q.
04

Analyzing Point R

Point R is at a height h and positioned horizontally at L. Because the box is accelerating horizontally, the fluid experiences an additional pressure due to this acceleration, which is \(\rho a L\). The total gauge pressure at R is the sum of the pressure due to the liquid above and due to the acceleration: \(P_{R} = \rho g h + \rho a L\).
05

Analyzing Point S

Point S is not at the bottom but horizontally opposite. The horizontal acceleration will induce a pressure at this point which is only due to the horizontal acceleration. Therefore, the gauge pressure at point S is \(P_{S} = \rho a L\).
06

Determine the Correct Option

By evaluating the gauge pressures at each point, we find: \(P_{P} = 0\), \(P_{Q} = \rho g h\), \(P_{R} = \rho g h + \rho a L\), and \(P_{S} = \rho a L\). This matches option (c).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Fluid Mechanics
Fluid mechanics is the branch of physics that studies the behavior of fluids (liquids, gases, and plasmas) and the forces on them. It is crucial for understanding how substances flow and how they interact with their surroundings. In this exercise, we deal specifically with a liquid in a rectangular box, which represents a typical problem in fluid mechanics.
  • The principles of fluid mechanics apply, including the use of pressure calculations and considerations of forces within a fluid.
  • Understanding how fluids behave under various conditions can help solve complex problems, such as those involving fluid flow and pressure variations.
This exercise involves applying the concept of gauge pressure in a dynamic scenario with horizontal acceleration. Understanding these fundamentals is key to deciphering the complex interplay of forces acting on the fluid.
Horizontal Acceleration Effects
In this scenario, the rectangular box filled with liquid is accelerating horizontally. This introduces additional forces and pressure variations within the fluid. Understanding the impact of horizontal acceleration on fluid dynamics is essential for solving problems like this one.
  • When the box accelerates, the fluid inside experiences a force due to the acceleration, impacting the gauge pressures at different points.
  • Points within the fluid that are to the right (or in the direction of acceleration) experience additional pressure increments, described as \( \rho a L \).
Considering horizontal acceleration effects helps predict and calculate changes in pressure at various locations within the liquid.
Density in Liquids
Density, symbolized as \( \rho \), is a central concept in fluid mechanics. It refers to the mass per unit volume of a substance, and in this problem, it plays a crucial role in determining pressure differences within the liquid.
  • Density determines how much weight a column of fluid exerts on points below it. This is critical in calculating gauge pressures at different depths.
  • The heavier (or denser) the liquid, the greater the pressure it will exert at a given depth or due to acceleration.
By understanding the role of density, we can better grasp how other factors like gravity and acceleration influence fluid behavior and pressure.
Pressure Calculation
Pressure calculation in this exercise involves both gravity-induced and acceleration-induced pressures. Gauge pressure is a measure of pressure relative to atmospheric pressure. Here's a breakdown of the pressure calculations in our problem:
  • At any point in the liquid, the pressure exerted by the fluid column above is given by \( \rho g h \), where \( h \) represents the fluid depth.
  • Horizontal acceleration introduces an additional pressure component: \( \rho a L \), adding to the pressure experienced horizontally.
By calculating the combined effects of these pressures, the true gauge pressure at specific points within the liquid can be effectively determined.
Rectangular Box Fluid Dynamics
When considering fluid dynamics within a rectangular box, the shape and orientation of the box play a significant role in determining how fluids behave under different forces. Understanding the fluid's reactions under these conditions is essential.
  • The shape of the box can affect how fluid moves and behaves under acceleration, influencing pressure distribution.
  • The pressure on a point within the box depends on the vertical height of the fluid above and the horizontal distance in direction of acceleration.
Predicting pressure changes involves understanding how fluids interact with their containers and the external forces acting upon them. This involves combining knowledge of fluid mechanics, pressure calculations, and the effects of acceleration.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A piece of wood is floating in water. When the temperature of water rises, the apparent weight of the wood will (a) increase (b) decrease (c) may increases or decrease (d) remain same

The working of venturimeter is based on (a) Torricelli's law (b) Pascal's law (c) Bernoulli's theorem (d) Archimedes principle

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

A cylindrical vessel contains a liquid of density \(\rho\) upto a height \(h\). The liquid is closed by a piston of mass \(m\) and area of cross-section \(A\). There is a small hole at the bottom of the vessel. The speed \(v\) with which the liquid comes out of the hole is (a) \(\sqrt{2 g h}\) (b) \(\sqrt{2\left(g h+\frac{m g}{\rho A}\right)}\) (c) \(\sqrt{2\left(g h+\frac{m g}{A}\right)}\) (d) \(\sqrt{2 g h+\frac{m g}{A}}\)

Two capillaries of length \(L\) and \(2 L\) and of radius \(R\) and \(2 R\) are connected in series. The net rate of flow of fluid through them will be (Given rate of the flow through single capillary, \(\left.X=\frac{\pi P R^{2}}{8 \eta L}\right)\) (a) \(\frac{8}{9} X\) (b) \(\frac{9}{8} X\) (c) \(\frac{5}{7} X\) (d) \(\frac{7}{5} X\)
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Astrophysics

Read Explanation

Electrostatics

Read Explanation

Magnetism

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.