Further Mathematics · AQA 7367 · Option M · Topic MA

Dimensional analysis

Checking that formulae make sense using dimensions (mass, length, time), and predicting formulae from them.

  • 2 sub-topics
  • Taught in Year 12
  • 22 practice questions
  • 2 animated lessons
Practice questions (22) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. MA1Dimensions of quantities and dimensional consistencyY12
  2. MA2Predicting formulae using dimensionsY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 22 questions in this topic, free, no account needed.

Dimensional analysis

Easy

Dimensions · Year 12 · 4 marks

  1. (a)
    Find the dimensions of pressure.
    (2)
  2. (b)
    Determine whether the equation P=FvP = Fv is dimensionally consistent.
    (2)
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Part (a)2 marks

  1. Write it in terms of mass, length and time.
  2. ML−1T−2ML^{-1}T^{-2}

Part (b)2 marks

  1. Find the dimensions of each term.
  2. Consistent: every term has the same dimensions.

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Dimensional analysis

Medium

Predicting formulae using dimensions · Year 12 · 5 marks

The speed vv of a wave on a string depends on the tension FF and the mass per unit length μ\mu, so that v=kFαμβv = kF^\alpha\mu^\beta.
  1. Use dimensional analysis to find the values of the indices.
    (5)
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  1. Write the dimensions of each quantity.
  2. Compare the powers of MM, LL and TT.M:0M: 0=α+β, L:1{}= \alpha + \beta, \ L: 1=α−β, T:−1{}= \alpha - \beta, \ T: -1=−2α{}= -2\alpha
  3. α\alpha=12, β{}= \tfrac{1}{2}, \ \beta=−12{}= -\tfrac{1}{2}

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Dimensional analysis

Hard

Predicting formulae · Year 12 · 3 marks

The period TT of a pendulum depends on length ll, mass mm and gg: T=klαmβgγT = k l^\alpha m^\beta g^\gamma.
  1. Find α\alpha, β\beta, γ\gamma.
    (3)

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  1. T\mathrm{T}=LαMβ(LT−2)γ{}= \mathrm{L}^{\alpha}\mathrm{M}^{\beta}(\mathrm{LT^{-2}})^{\gamma}
  2. β\beta=0, γ{}= 0, \ \gamma=−12{}= -\tfrac{1}{2}
  3. α=12\alpha = \tfrac{1}{2}

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