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Units and Measurements - Dimensions of Physical Quantities

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 base quantities are raised to represent that quantity. The seven base quantities are Mass [M][M], Length [L][L], Time [T][T], Electric Current [A][A], Thermodynamic Temperature [K][K], Amount of Substance [mol][mol], and Luminous Intensity [cd][cd].

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A dimensional formula is an expression showing how and which of the base quantities represent the dimensions of a physical quantity, typically written as [MaLbTc][M^a L^b T^c].

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The Principle of Homogeneity of Dimensions states that a physical equation is dimensionally correct only if the dimensions of all the terms on both sides of the equation are the same.

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Applications of Dimensional Analysis include: (1) Checking the dimensional consistency of equations, (2) Converting units from one system to another using the relation n1u1=n2u2n_1 u_1 = n_2 u_2, and (3) Deriving relations between various physical quantities.

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Limitations of Dimensional Analysis: It 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

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

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

[Density]=[ML−3][Density] = [ML^{-3}]

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

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

[Force]=[MLT−2][Force] = [MLT^{-2}]

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

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

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

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

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:

Find the dimensional formula of the Universal Gravitational Constant GG using the formula F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}.

Solution:

Rearranging for GG: G=Fr2m1m2G = \frac{F r^2}{m_1 m_2} Substituting dimensions of Force [F]=[MLT−2][F] = [MLT^{-2}], distance [r]=[L][r] = [L], and mass [m]=[M][m] = [M]: [G]=[MLT−2][L2][M][M]=[ML3T−2][M2]=[M−1L3T−2][G] = \frac{[MLT^{-2}][L^2]}{[M][M]} = \frac{[ML^3T^{-2}]}{[M^2]} = [M^{-1}L^3T^{-2}]

Explanation:

By isolating the constant GG and substituting the known dimensions of Force, length, and mass, we derive the dimensions for GG as [M−1L3T−2][M^{-1}L^3T^{-2}].

Problem 2:

Check the dimensional correctness 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: [s]=[L][s] = [L] Dimensions of RHS terms: [ut]=[LT−1][T]=[L][ut] = [LT^{-1}][T] = [L] [12at2]=[LT−2][T2]=[L][\frac{1}{2}at^2] = [LT^{-2}][T^2] = [L] Since dimensions of LHS = dimensions of RHS for each term, the equation is dimensionally correct.

Explanation:

According to the Principle of Homogeneity, each term added or subtracted in an equation must have the same dimensions. Here, all terms have the dimension [L][L].

Problem 3:

In the equation v=a+btv = a + bt, where vv is velocity and tt is time, find the dimensions of constants aa and bb.

Solution:

By the Principle of Homogeneity: [a]=[v]=[LT−1][a] = [v] = [LT^{-1}] and [bt]=[v]  ⟹  [b][T]=[LT−1][bt] = [v] \implies [b][T] = [LT^{-1}] [b]=[LT−1][T]=[LT−2][b] = \frac{[LT^{-1}]}{[T]} = [LT^{-2}]

Explanation:

The dimensions of each term in the sum must equal the dimensions of the quantity on the left side (vv). Therefore, aa has dimensions of velocity and bb has dimensions of acceleration.