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Units and Measurements - Dimensional Formulae and Dimensional Equations

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Dimensions of a physical quantity are the powers to which the fundamental units (Mass [M][M], Length [L][L], Time [T][T], etc.) are raised to represent that quantity.

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A Dimensional Formula is an expression showing which of the base quantities and with what powers they enter into the derived unit of a physical quantity, written as [MaLbTc][M^a L^b T^c].

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A Dimensional Equation is obtained by equating a physical quantity with its dimensional formula, e.g., [Force]=[MLT−2][Force] = [M L T^{-2}].

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The Principle of Homogeneity of Dimensions states that the dimensions of each term of a physical equation must be the same on both sides. This allows us to check the correctness of an equation: only quantities with the same dimensions can be added or subtracted.

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Applications of Dimensional Analysis: (i) Checking the dimensional consistency of equations, (ii) Deducing relations among physical quantities, and (iii) Conversion of units from one system to another using the formula n1u1=n2u2n_1 u_1 = n_2 u_2.

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Limitations: Dimensional analysis cannot determine dimensionless constants, it fails if a quantity depends on more than three fundamental quantities (in mechanics), and it cannot derive equations involving trigonometric, logarithmic, or exponential functions.

📐Formulae

[Area]=[L2][Area] = [L^2]

[Volume]=[L3][Volume] = [L^3]

[Velocity]=[LT−1][Velocity] = [L T^{-1}]

[Acceleration]=[LT−2][Acceleration] = [L T^{-2}]

[Force]=[MLT−2][Force] = [M L T^{-2}]

[Work]=[Energy]=[ML2T−2][Work] = [Energy] = [M L^2 T^{-2}]

[Power]=[ML2T−3][Power] = [M L^2 T^{-3}]

[Pressure]=[Stress]=[ML−1T−2][Pressure] = [Stress] = [M L^{-1} T^{-2}]

[Universal Gravitational Constant (G)]=[M−1L3T−2][Universal\, Gravitational\, Constant\, (G)] = [M^{-1} L^3 T^{-2}]

[Planck′s Constant (h)]=[ML2T−1][Planck's\, Constant\, (h)] = [M L^2 T^{-1}]

💡Examples

Problem 1:

Check the dimensional consistency of the equation s=ut+12at2s = ut + \frac{1}{2}at^2, where ss is displacement, uu is initial velocity, aa is acceleration, and tt is time.

Solution:

Dimensions of LHS (displacement ss): [L][L]. \nDimensions of terms on RHS:

  1. [ut]=[LT−1][T]=[L][ut] = [L T^{-1}][T] = [L]
  2. [12at2]=[LT−2][T2]=[L][\frac{1}{2}at^2] = [L T^{-2}][T^2] = [L] (Note: 12\frac{1}{2} is a dimensionless constant).

Explanation:

Since the dimensions of all terms on both the LHS and RHS are [L][L], the equation is dimensionally correct according to the Principle of Homogeneity.

Problem 2:

In the van der Waals equation (P+aV2)(V−b)=RT(P + \frac{a}{V^2})(V - b) = RT, find the dimensions of the constants aa and bb, where PP is pressure and VV is volume.

Solution:

According to the Principle of Homogeneity:

  1. bb can only be subtracted from VV if they have the same dimensions. So, [b]=[V]=[L3][b] = [V] = [L^3].
  2. aV2\frac{a}{V^2} can only be added to PP if they have the same dimensions. [a/V2]=[P]  ⟹  [a]=[P][V2][a/V^2] = [P] \implies [a] = [P][V^2] [a]=[ML−1T−2][L3]2=[ML−1T−2][L6]=[ML5T−2][a] = [M L^{-1} T^{-2}][L^3]^2 = [M L^{-1} T^{-2}][L^6] = [M L^5 T^{-2}].

Explanation:

The constants aa and bb must have dimensions such that the terms being added or subtracted are dimensionally identical.

Problem 3:

Calculate the dimensions of Torque.

Solution:

Torque (τ)=Force×perpendicular distance(\tau) = Force \times perpendicular\, distance [τ]=[MLT−2]×[L]=[ML2T−2][\tau] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}] \nThis is the same dimensional formula as Work and Energy.

Explanation:

Torque is a rotational analog of force, and its dimensional formula highlights its relationship with energy (though it is a vector quantity).