Dimensional Analysis Problems
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Questions and Answers

Which of the following factors influences the pressure variation per unit length in a pipe?

  • Diameter of the pipe
  • Temperature of the water
  • Length of the pipe
  • Weight of the water (correct)
  • If the expression $Sen heta = rac{PRx + QBz}{m}$ is dimensionally homogeneous, which dimension corresponds to Q?

  • M^2LT^{-1} (correct)
  • MT^{-2}
  • MLT
  • M^2L^{-2}T^{-1}
  • In the equation $V = rac{3V^2aFy}{Sen(zay)} - xF$, what dimensions must the variable z have?

  • LT^{-1} (correct)
  • M^0L^0T^0
  • ML^2T^{-2}
  • M^2LT^{-1}
  • For the expression $ rac{e^{-Pvt}}{D_0}$ to be dimensionally correct, what condition must be satisfied by P?

    <p>P must have dimensions of MLT^{-2}</p> Signup and view all the answers

    What are the dimensions of K in the equation $K^2 = rac{F}{6 ext{√}(PD^2V^{-1})}$?

    <p>M^2L^{-1}T^{-2}</p> Signup and view all the answers

    Which of the following is the unit of magnetic permeability μ in the SI system?

    <p>H·m^{-1}</p> Signup and view all the answers

    What is the dimensional equation of X in the equation $ rac{F}{V} = rac{9.8 * P * ext{√}5 * m * sen(37°) }{X}$?

    <p>M^1L^3T^{-2}</p> Signup and view all the answers

    In a new unit system where velocity, mass, and force are fundamental, what are the dimensions of electric potential energy E?

    <p>B^2C^{-1}</p> Signup and view all the answers

    Study Notes

    Dimensional Analysis - Problem 1

    • The pressure per unit of length depends on the weight of the water through the pipe, the speed of the water, and the acceleration of gravity.
    • These factors are related by a constant, k, which is determined experimentally.

    Dimensional Analysis - Problem 2

    • To ensure dimensional homogeneity, the dimensions of "Q" must be determined.
    • The expression involves the following variables:
      • m = mass
      • R = radius
      • x = time
      • B = force
      • Z = velocity
    • The dimensions of "Q" can be calculated by analyzing the dimensions of each variable.

    Dimensional Analysis - Problem 3

    • The equation for a physical phenomenon is given.
    • The equation involves velocity (V), force (F), and acceleration (a).
    • The dimensions of "z" need to be determined.

    Dimensional Analysis - Problem 4

    • The expression is: $\sum_{i=1}^{n}D_1ce= \frac{e^{-Pvt}}{D_0}$
    • The variables involved are:
      • v = linear velocity
      • D0, D1 = Density
      • c, e = length
      • t = time
    • The value of "P" needs to be chosen to ensure that the expression is dimensionally correct.

    Dimensional Analysis - Problem 5

    • The equation is: K2=F6PD2V−1K^2 = \frac{F}{6\sqrt{PD^2V^{-1}}}K2=6PD2V−1​F​
    • The variables involved are:
      • F = Force
      • P = Pressure
      • D = Density
      • v = Velocity.
    • The dimensional equation for "K" needs to be determined.

    Dimensional Analysis - Problem 6

    • The density of the magnetic flux "B" is given by: B=μI2πrB = \frac{\mu I}{2\pi r}B=2πrμI​
    • The variables involved are:
      • B = Magnetic flux density
      • I = Current
      • μ = Magnetic permeability
      • r = Radial distance
    • The unit of the magnetic permeability "μ" needs to be determined in the SI unit system.

    Dimensional Analysis - Problem 7

    • The equation is: FV=9,8∗P∗5∗m∗sen(37°)X\frac{F}{V} = \frac{9,8 * P * \sqrt{5} * m * sen(37°) }{X}VF​=X9,8∗P∗5​∗m∗sen(37°)​
    • The variables involved are:
      • F = Force
      • V = Volume
      • P = Power
      • m = Length
      • X = Unknown variable
    • The dimensional equation for "X" needs to be determined.

    Dimensional Analysis - Problem 8

    • The critical velocity "v" of a liquid flowing through a pipe depends on:
      • viscosity "η"
      • density "ρ"
      • diameter "D"
      • a dimensionless constant "R"
    • The dimensions of "v" need to be expressed in terms of the dimensions of "η", "ρ", "D", and "R".

    Dimensional Analysis - Problem 9

    • A new unit system is defined where velocity, mass, and force are the fundamental magnitudes.
    • The new system uses A for velocity, B for mass, and C for force.
    • The dimensional equation of "E" needs to be determined in this new system, knowing that E = pressure x density.

    Dimensional Analysis - Problem 10

    • The electric potential energy (UE) is given by: UE=kQ2dU_E = \frac {kQ^2}{d}UE​=dkQ2​
    • The variables are:
      • Q = Electric charge
      • d = Distance.
    • The dimensions of "k" needs to be determined.

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    Description

    This quiz focuses on various problems related to dimensional analysis. Each problem involves factors like mass, force, velocity, and their relationships. Test your understanding of how to analyze dimensions and maintain dimensional homogeneity in physical equations.

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