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Units and Measurements - Dimensional Analysis and its Applications

Grade 11CBSEPhysics

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

🔑Concepts

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Dimensions of a physical quantity are the powers to which the base quantities are raised to represent that quantity. For example, Force is [MLT−2][M L T^{-2}].

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Dimensional Formula is an expression showing which of the fundamental quantities and with what powers enter into the derived unit of a physical quantity.

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The Principle of Homogeneity of Dimensions states that a physical equation is dimensionally correct if the dimensions of all the terms on both sides of the equation are the same. Only quantities with the same dimensions can be added, subtracted, or compared.

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Application 1: Checking the Dimensional Correctness of an Equation. If the dimensions of LHS≠RHSLHS \neq RHS, the equation is wrong.

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Application 2: Deducing Relations among Physical Quantities. By assuming a quantity depends on powers of other quantities, we can find the functional relationship using dimensional balance.

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Application 3: Conversion of Units. Changing the magnitude of a physical quantity from one system of units to another using the fact that n[u]n[u] is constant, where nn is the numerical value and uu is the unit.

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

📐Formulae

[PhysicalQuantity]=[MaLbTc][Physical Quantity] = [M^a L^b T^c]

n2=n1[M1M2]a[L1L2]b[T1T2]cn_2 = n_1 \left[ \frac{M_1}{M_2} \right]^a \left[ \frac{L_1}{L_2} \right]^b \left[ \frac{T_1}{T_2} \right]^c

Work done=[M1L2T−2]\text{Work done} = [M^1 L^2 T^{-2}]

Pressure=[Force][Area]=[ML−1T−2]\text{Pressure} = \frac{[Force]}{[Area]} = [M L^{-1} T^{-2}]

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

💡Examples

Problem 1:

Check the dimensional correctness of the equation v2−u2=2asv^2 - u^2 = 2as, where vv is final velocity, uu is initial velocity, aa is acceleration, and ss is displacement.

Solution:

Dimensions of LHS: [v2]=[LT−1]2=[L2T−2][v^2] = [LT^{-1}]^2 = [L^2 T^{-2}] and [u2]=[LT−1]2=[L2T−2][u^2] = [LT^{-1}]^2 = [L^2 T^{-2}]. Thus, LHS = [L2T−2][L^2 T^{-2}]. Dimensions of RHS: [2as]=[1][LT−2][L]=[L2T−2][2as] = [1][LT^{-2}][L] = [L^2 T^{-2}].

Explanation:

Since the dimensions of each term on the LHS are equal to the dimensions of the term on the RHS, the equation is dimensionally correct according to the Principle of Homogeneity.

Problem 2:

Derive an expression for the time period TT of a simple pendulum which may depend on mass of the bob mm, length of the pendulum ll, and acceleration due to gravity gg.

Solution:

Let T∝malbgcT \propto m^a l^b g^c. Writing dimensions: [M0L0T1]=[M]a[L]b[LT−2]c[M^0 L^0 T^1] = [M]^a [L]^b [LT^{-2}]^c. Equating powers: a=0a = 0 (mass), b+c=0b + c = 0 (length), −2c=1-2c = 1 (time). Solving gives c=−1/2c = -1/2, b=1/2b = 1/2, a=0a = 0. So, T=klgT = k \sqrt{\frac{l}{g}}.

Explanation:

By equating the exponents of M,L,M, L, and TT on both sides, we find that the time period is independent of mass and proportional to the square root of the ratio of length to gravity.

Problem 3:

Find the dimensions of constants aa and bb in the Van der Waals equation: (P+aV2)(V−b)=RT\left(P + \frac{a}{V^2}\right)(V - b) = RT.

Solution:

By Principle of Homogeneity, [aV2]=[P]\left[\frac{a}{V^2}\right] = [P]. Thus [a]=[P][V2]=[ML−1T−2][L3]2=[ML5T−2][a] = [P][V^2] = [ML^{-1}T^{-2}][L^3]^2 = [ML^5T^{-2}]. Similarly, [b]=[V]=[L3][b] = [V] = [L^3].

Explanation:

Quantities added to or subtracted from each other must have identical dimensions. Therefore, a/V2a/V^2 must have dimensions of Pressure (PP) and bb must have dimensions of Volume (VV).

Dimensional Analysis and its Applications Revision - Class 11 Physics CBSE