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Monday, 9 June 2025

Worksheet: Dimensional Formulae and Dimensional Equations


📝 Worksheet: Dimensional Formulae and Dimensional Equations


✍️ Section A: Fill in the Blanks

  1. The dimensional formula of force is ________________.

  2. The dimensional formula of energy is ________________.

  3. The dimensional formula of pressure is ________________.

  4. The dimensional formula of power is ________________.

  5. The dimensional formula of momentum is ________________.


📘 Section B: Match the Following

Physical Quantity Dimensional Formula
A. Work/Energy i. [M1L1T2][M^1 L^1 T^{-2}]
B. Acceleration ii. [M1L2T3][M^1 L^2 T^{-3}]
C. Power iii. [M1L2T2][M^1 L^2 T^{-2}]
D. Force iv. [M0LT2][M^0 L T^{-2}]
E. Velocity v. [M0LT1][M^0 L T^{-1}]


🧠 Section C: Conceptual Questions

  1. What is meant by the dimensional formula of a physical quantity?

  2. How can you check whether a physical equation is dimensionally correct?

  3. Why can’t dimensional analysis help in finding constants like 12\frac{1}{2}, π\pi, or 2?

  4. Can two different physical quantities have the same dimensional formula? Give an example.

  5. Why is dimensional analysis called a “preliminary test” for checking the correctness of equations?


🔎 Section D: Derivation Questions

Find the dimensional formula of the following quantities by analyzing their base physical relation.

  1. Kinetic Energy: KE=12mv2KE = \frac{1}{2}mv^2

  2. Pressure: P=FAP = \frac{F}{A}

  3. Momentum: p=mvp = mv

  4. Gravitational Potential Energy: U=mghU = mgh

  5. Surface Tension: T=FlT = \frac{F}{l}


🔍 Section E: Dimensional Consistency Check

Check whether the following equations are dimensionally correct.

  1. v=u+atv = u + at

  2. s=ut+12at2s = ut + \frac{1}{2}at^2

  3. E=mc3E = mc^3

  4. T=2πlgT = 2\pi \sqrt{\frac{l}{g}} (Time period of a simple pendulum)

  5. F=ma+vF = ma + v


📐 Section F: Assertion and Reasoning (Write TRUE/FALSE)

  1. Assertion: The dimensional formula of work is the same as that of energy.
    Reason: Both involve force applied over a distance.

  2. Assertion: Dimensional equations are unique for all physical quantities.
    Reason: Each physical quantity is derived from different base units.



Answer Key


✍️ Section A: Fill in the Blanks

  1. [M1L1T2][M^1 L^1 T^{-2}]

  2. [M1L2T2][M^1 L^2 T^{-2}]

  3. [M1L1T2][M^1 L^{-1} T^{-2}]

  4. [M1L2T3][M^1 L^2 T^{-3}]

  5. [M1L1T1][M^1 L^1 T^{-1}]


📘 Section B: Match the Following

Physical Quantity Dimensional Formula
A. Work/Energy iii. [M1L2T2][M^1 L^2 T^{-2}]
B. Acceleration iv. [M0LT2][M^0 L T^{-2}]
C. Power ii. [M1L2T3][M^1 L^2 T^{-3}]
D. Force i. [M1L1T2][M^1 L^1 T^{-2}]
E. Velocity v. [M0LT1][M^0 L T^{-1}]


🧠 Section C: Conceptual Questions

  1. Dimensional formula expresses a physical quantity in terms of the base quantities (M, L, T, etc.) with their respective powers.

  2. By checking whether both sides of the equation have the same dimensional formula.

  3. Because dimensional analysis only handles dimensions, not pure numbers or constants without dimension.

  4. Yes. Torque and work both have the same dimensional formula: [ML2T2][M L^2 T^{-2}].

  5. Because it only verifies dimensional homogeneity, not correctness of constants or functional dependence.


🔎 Section D: Derivation Questions

  1. KE=12mv2KE = \frac{1}{2}mv^2:
    [M][LT1]2=[ML2T2][M] \cdot [L T^{-1}]^2 = [M L^2 T^{-2}]

  2. P=FAP = \frac{F}{A}:
    [MLT2]/[L2]=[ML1T2][M L T^{-2}]/[L^2] = [M L^{-1} T^{-2}]

  3. p=mvp = mv:
    [M][LT1]=[MLT1][M] \cdot [L T^{-1}] = [M L T^{-1}]

  4. U=mghU = mgh:
    [M][LT2][L]=[ML2T2][M] \cdot [L T^{-2}] \cdot [L] = [M L^2 T^{-2}]

  5. T=FlT = \frac{F}{l}:
    [MLT2]/[L]=[MT2][M L T^{-2}]/[L] = [M T^{-2}]


🔍 Section E: Dimensional Consistency Check

  1. v=u+atv = u + at
    Correct (All terms have [LT1][L T^{-1}])

  2. s=ut+12at2s = ut + \frac{1}{2}at^2
    Correct ([L][L] on all terms)

  3. E=mc3E = mc^3
    Incorrect
    LHS: [ML2T2][M L^2 T^{-2}]
    RHS: [M][LT1]3=[ML3T3][M] \cdot [L T^{-1}]^3 = [M L^3 T^{-3}]

  4. T=2πlgT = 2\pi \sqrt{\frac{l}{g}}
    Correct
    [L]/[LT2]=[T2]T2=[T][L]/[L T^{-2}] = [T^2] \Rightarrow \sqrt{T^2} = [T]

  5. F=ma+vF = ma + v
    Incorrect
    LHS: [MLT2][M L T^{-2}], RHS: sum of force and velocity → dimensionally incompatible


📐 Section F: Assertion and Reasoning

  1. TRUE – Both involve application of force over distance and have the same dimensional formula.

  2. FALSE – Different quantities can share the same dimensional formula (e.g., torque and work).


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