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Chapter 1: Problem 38

The largest mass \((m)\) that can be moved by a flowing river depends on velocity \((v)\), density \((\rho)\) of river water and acceleration due to gravity \((g)\). The correct relation is (a) \(m \propto \frac{\rho^{2} v^{4}}{g^{2}}\) (b) \(m \propto \frac{\rho v^{6}}{g^{2}}\) (c) \(m \propto \frac{\rho v^{4}}{g^{3}}\) (d) \(m \propto \frac{\rho v^{6}}{g^{3}}\)

Short Answer

Expert verified
The correct relation is (a) \(m \propto \frac{\rho^{2} v^{4}}{g^{2}}\)

Step by step solution

01

Identify the dimensions of each variable

First, identify the dimensions of the variables in the problem. The mass \((m)\) is measured in kilograms \((kg)\). The velocity \((v)\) is measured in meters per second \((m/s)\). The density \((\rho)\) is measured in kilograms per cubic meter \((kg/m^3)\). The acceleration due to gravity \((g)\) is measured in meters per second squared \((m/s^2)\).
02

Analyze each option

Now look at each option and determine which would result in an expression with the same dimensions as mass (kg). With the given dimensions, the only option which will have the dimensions as of mass is (a) \(m \propto \frac{\rho^{2} v^{4}}{g^{2}}\) .
03

Confirm the answer

To confirm the validity of this answer, substitute the dimensions of each of the quantities into the chosen option (a). After simplification, it should be clear that the resulting dimensional formula is indeed mass.

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Key Concepts

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

Physics Problem Solving
Physics problem-solving is an essential skill that involves analyzing a situation to determine how physical laws and principles apply. For this specific problem, we focus on finding a relationship between the mass moved by a river and other physical quantities like velocity, density, and gravity.

When solving physics problems, it's crucial to:
  • Understand the problem by identifying knowns and unknowns. Here, the largest mass that can be moved is influenced by velocity, density, and acceleration due to gravity.
  • Use dimensional analysis as a powerful tool. It helps confirm whether a derived relation is dimensionally consistent.
  • Systematically break down complex relationships into simpler parts. In this problem, by examining how each variable's dimensions contribute to the final dimension of mass, students can identify the correct relationship among the given options.
With careful analysis and a structured approach, students can navigate through physics problems effectively. Dimensional analysis ensures consistency and provides a logic check for equations.
Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In the context of this problem, the velocity (\(v\)) of the river is a key kinematic variable that determines the movement of mass.

Velocity, a fundamental concept in kinematics, measures how quickly an object changes its position. In our problem:
  • Velocity is directly related to the energy and ability of the river to move mass.
  • The higher the velocity, the greater the potential to transport more massive objects.
  • The kinematic equation accommodates for velocity's contribution in relation to other variables, ensuring a coherent understanding of its role in mass transport in the river.
Understanding kinematics is crucial in solving physics problems where motion plays a central role, as it often does in fluid dynamics and mechanics.
Fluid Mechanics
Fluid mechanics examines how fluids (liquids and gases) behave when at rest and in motion. This problem involves river water, emphasizing how fluid properties impact physical scenarios like moving mass.

Some key aspects of fluid mechanics seen here include:
  • Density (\(\rho\)): This property affects the buoyancy and mass transportation capabilities of the fluid鈥攊n this case, the river water.
  • Physical interaction: Fluid mechanics explains how intermolecular forces and fluidity allow the river to transfer energy and move objects of given mass.
  • Flow dynamics: As the river flows with a particular velocity, these dynamics determine how much energy can be used to transport mass are crucial.
By understanding these concepts, students can better grasp how rivers can move various masses and the factors that impact this ability. Fluid mechanics provides a framework for analyzing such environmental processes and systems.

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Most popular questions from this chapter

The volume \(V\) of water passing through a point of a uniform tube during \(t\) seconds is related to the crosssectional area \(A\) of the tube and velocity \(u\) of water by the relation \(V \propto A^{\alpha} u^{\beta} t^{\gamma}\), which one of the following will be true (a) \(\alpha=\beta=\gamma\) (b) \(\alpha \neq \beta=\gamma\) (c) \(\alpha=\beta \neq \gamma\) (d) \(\alpha \neq \beta \neq \gamma\)

If \(P\) represents radiation pressure, \(C\) represents speed of light and \(Q\) represents radiation energy striking a unit area per second, then non-zero integers \(x, y\) and \(z\) such that \(P^{x} Q^{y} C^{z}\) is dimensionless, are (a) \(x=1, y=1, z=-1\) (b) \(x=1, y=-1, z=1\) (c) \(x=-1, y=1, z=1\) (d) \(x=1, y=1, z=1\)

If there is a positive error of \(50 \%\) in the measurement of velocity of a body, then the error in the measurement of kinetic energy is (a) \(25 \%\) (b) \(50 \%\) (c) \(100 \%\) (d) \(125 \%\)

In two systems of relations among velocity, acceleration and force are respectively \(v_{2}=\frac{\alpha^{2}}{\beta} v_{1}\), \(a_{2}=\alpha \beta a_{1}\) and \(F_{2}=\frac{F_{1}}{\alpha \beta}\). If \(\alpha\) and \(\beta\) are constants then relations among mass, length and time in two systems are (a) \(M_{2}=\frac{\alpha}{\beta} M_{1}, L_{2}=\frac{\alpha^{2}}{\beta^{2}} L_{1}, T_{2}=\frac{\alpha^{3} T_{1}}{\beta}\) (b) \(M_{2}=\frac{1}{\alpha^{2} \beta^{2}} M_{1}, L_{2}=\frac{\alpha^{3}}{\beta^{3}} L_{1}, T_{2}=T_{1} \frac{\alpha}{\beta^{2}}\) (c) \(M_{2}=\frac{\alpha^{3}}{\beta^{3}} M_{1}, L_{2}=\frac{\alpha^{2}}{\beta^{2}} L_{1}, T_{2}=\frac{\alpha}{\beta} T_{1}\) (d) \(M_{2}=\frac{\alpha^{2}}{\beta^{2}} M_{1}, L_{2}=\frac{\alpha}{\beta^{2}} L_{1}, T_{2}=\frac{\alpha^{3}}{\beta^{3}} T_{1}\)

The potential energy of a particle varies with distance \(x\) from a fixed origin as \(U=\frac{A \sqrt{x}}{x^{2}+B}\), where \(A\) and \(B\) are dimensional constants then dimensional formula for \(A B\) is (a) \(M L^{7 / 2} T^{-2}\) (b) \(M L^{11 / 2} T^{-2}\) (c) \(M^{2} L^{9 / 2} T^{-2}\) (d) \(M L^{13 / 2} T^{-3}\)
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