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PhysicsMediumClass 11

Dimensional Analysis — Deriving relations between quantities

Units & Measurements

10

JEE Qs

8%

Hard

60

min

Master the dimensions of common physical quantities and practice solving systems of linear equations to accurately determine the powers in derived relations.

🧮 Key Formulas

Principle of Dimensional Homogeneity: [LHS] = [RHS] (dimensions of all terms in a physical equation must be the same)
Assumed relation form: Q = k * A^a * B^b * C^c (where Q is the dependent quantity, A, B, C are independent quantities, k is a dimensionless constant, and a, b, c are powers)

✅ Key Points for JEE

  • 1The principle of dimensional homogeneity is the core concept: dimensions of all terms on both sides of a physically correct equation must be identical.
  • 2To derive a relation, assume the dependent quantity is proportional to a product of powers of the independent quantities (e.g., T = k * l^a * g^b).
  • 3Equate the powers of fundamental dimensions (M, L, T, etc.) on both sides of the dimensionally homogeneous equation to form a system of linear equations.
  • 4Solve the system of linear equations to find the values of the powers (a, b, c).
  • 5Dimensional analysis cannot determine dimensionless constants, nor can it derive relations involving sums/differences of quantities with different dimensions, or trigonometric/exponential/logarithmic functions.

⚠️ Common Mistakes

  • Incorrectly identifying or calculating the dimensional formulas of the physical quantities involved.
  • Errors in setting up or solving the simultaneous linear equations for the powers of the fundamental dimensions.
  • Forgetting that dimensional analysis only provides the proportionality of quantities, not the exact numerical constant.
  • Attempting to apply dimensional analysis where the number of independent variables is greater than the number of fundamental dimensions, leading to underdetermined systems.

📝 Practice Questions

See all

Q43.The maximum percentage error in the measurment of density of a wire is [Given, mass of wire = (0.60 ± 0.003)g radius of wire = (0.50 ± 0.01)cm length of wire = (10.00 ± 0.05)cm] (1) 8 (2) 5 (3) 4 (4) 7 for another diatomic molecules, but for rigid molecules and γ2 =

2025·MCQMedium

Q43.The energy of a system is given as E(t) = α3e−βt , where t is the time and β = 0.3 s−1 . The errors in the measurement of α and t are 1.2% and 1.6% , respectively. At t = 5 s, maximum percentage error in the energy is : (1) 6% (2) 8.4% (3) 11.6% (4) 4%

2025·MCQMedium

Q39.Given below are two statements : Statement I : In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II : The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. In the light of the above statements, choose the correct answer from the options given below. (1) Statement I is true but Statement II is false (2) Statement I is false but Statement II is true (3) Both Statement I and Statement II are false (4) Both Statement I and Statement II are true 2025 (22 Jan Shift 1) JEE Main Previous Year Paper

2025·Assertion ReasoningMedium

Q40.Which one of the following is the correct dimensional formula for the capacitance in F ? M, L, T and C stand for unit of mass, length, time and charge, (1) [F] = [C2M−1 L−2 T2] (2) [F] = [C2M−2 L2 T2] (3) [F] = [CM−2 L−2 T−2] (4) [F] = [CM−1 L−2 T2]

2025·MCQMedium

Q28.The position of a particle moving on x-axis is given by x(t) = A sin t + B cos2 t + Ct2 + D , where t is time. The dimension of ABC is D (1) L2T −2 (2) L2 (3) L (4) L3 T−2

2025·MCQEasy

Q45.For an experimental expression y = 32.3×112527.4 , where all the digits are significant. Then to report the value of y we should write (1) y = 1326.19 (2) y = 1330 (3) y = 1326.186 (4) y = 1326.2

2025·MCQEasy

NCERT Chapters

  • Class 11 Physics Ch 2: Units and Measurements