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Physical World and Measurement - Dimensional Analysis

Grade 11ICSEPhysics

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 (M,L,TM, L, T, etc.) are raised to represent that quantity.

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The Dimensional Formula is an expression showing which of the fundamental quantities and with what powers enter into the derived unit of a physical quantity, e.g., Force is [M1L1T−2][M^1 L^1 T^{-2}].

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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. This is based on the fact that only physical quantities of the same kind can be added, subtracted, or compared.

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Uses of Dimensional Analysis: (1) To check the correctness of a physical equation. (2) To derive the relationship between various physical quantities. (3) To convert the value of a physical quantity from one system of units to another using the relation n1[u1]=n2[u2]n_1 [u_1] = n_2 [u_2].

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Limitations: Dimensional analysis cannot determine dimensionless constants (like π\pi or 22), cannot derive relations involving logarithmic, exponential, or trigonometric functions, and cannot be used if a physical quantity depends on more than three fundamental quantities (if only M,L,TM, L, T are involved).

📐Formulae

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

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

Force [F]=[MLT−2]\text{Force } [F] = [M L T^{-2}]

Work / Energy [W]=[ML2T−2]\text{Work / Energy } [W] = [M L^2 T^{-2}]

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

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

Planck’s Constant [h]=[ML2T−1]\text{Planck's Constant } [h] = [M L^2 T^{-1}]

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

💡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:

LHS: [v2]=[LT−1]2=[L2T−2][v^2] = [L T^{-1}]^2 = [L^2 T^{-2}] and [u2]=[LT−1]2=[L2T−2][u^2] = [L T^{-1}]^2 = [L^2 T^{-2}]. Since we subtract like quantities, the dimensions of LHS are [L2T−2][L^2 T^{-2}].

RHS: [2as]=[1][LT−2][L]=[L2T−2][2as] = [1][L T^{-2}][L] = [L^2 T^{-2}] (2 is a dimensionless constant).

Explanation:

Since the dimensions of the LHS ([L2T−2][L^2 T^{-2}]) are equal to the dimensions of the RHS ([L2T−2][L^2 T^{-2}]), 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 the 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 on both sides: [M0L0T1]=[M]a[L]b[LT−2]c[M^0 L^0 T^1] = [M]^a [L]^b [L T^{-2}]^c [M0L0T1]=[MaLb+cT−2c][M^0 L^0 T^1] = [M^a L^{b+c} T^{-2c}]. Equating powers: a=0a = 0 b+c=0b + c = 0 −2c=1  ⟹  c=−1/2-2c = 1 \implies c = -1/2. Substituting cc into b+c=0b + c = 0, we get b=1/2b = 1/2.

Explanation:

Substituting a,b,ca, b, c back into the original relation, we get T∝m0l1/2g−1/2T \propto m^0 l^{1/2} g^{-1/2}, which simplifies to T=klgT = k \sqrt{\frac{l}{g}}. Experimentally, k=2πk = 2\pi.

Problem 3:

Convert 1 Joule1 \text{ Joule} (SI unit of work) into erg\text{erg} (CGS unit) using dimensional analysis.

Solution:

Dimensions of Work [W]=[ML2T−2][W] = [M L^2 T^{-2}]. Here a=1,b=2,c=−2a=1, b=2, c=-2. In SI (n1=1n_1=1): M1=1 kg,L1=1 m,T1=1 sM_1 = 1\text{ kg}, L_1 = 1\text{ m}, T_1 = 1\text{ s}. In CGS: M2=1 g,L2=1 cm,T2=1 sM_2 = 1\text{ g}, L_2 = 1\text{ cm}, T_2 = 1\text{ s}. n2=1[1 kg1 g]1[1 m1 cm]2[1 s1 s]−2n_2 = 1 \left[\frac{1\text{ kg}}{1\text{ g}}\right]^1 \left[\frac{1\text{ m}}{1\text{ cm}}\right]^2 \left[\frac{1\text{ s}}{1\text{ s}}\right]^{-2} n2=1×[103]×[102]2×[1]−2=103×104=107n_2 = 1 \times [10^3] \times [10^2]^2 \times [1]^{-2} = 10^3 \times 10^4 = 10^7.

Explanation:

Thus, 1 Joule=107 erg1 \text{ Joule} = 10^7 \text{ erg}.

Dimensional Analysis - Revision Notes & Key Formulas | ICSE Class 11 Physics