JAMB 2018 Physics · Question 85

Which of the following is the dimension of pressure

Options

  1. A.ML−1^{−1}T−2^{−2}
  2. B.MLT−2^{−2}
  3. C.ML2^{2}T3^{3}
  4. D.ML−3^{−3}

Answer: A. ML−1^{−1}T−2^{−2}

Explanation written with AI and checked against the answer key

1. Analyze the question

The question asks for the dimensional formula of pressure. To find this, we need to break down pressure into its fundamental physical quantities: Mass (MM), Length (LL), and Time (TT).

2. Explain the concept

In physics, dimensions are expressed in terms of the base units of the SI system.

  • Pressure (PP) is defined as Force (FF) per unit Area (AA): $P=FAP = \frac{F}{A}$
  • Force (FF) is defined by Newton's Second Law as mass (mm) times acceleration (aa): $F=m×aF = m \times a$ • Dimension of mass (mm) = [M][M] • Dimension of acceleration (aa) = [LT−2][LT^{-2}] • Therefore, dimension of Force = [MLT−2][MLT^{-2}]
  • Area (AA) is length times width (or length squared): • Dimension of Area = [L2][L^2]
3. Show the solution

Now, substitute the dimensions of Force and Area into the pressure formula:

[P]=[Force][Area][P] = \frac{[Force]}{[Area]}
[P]=[MLT−2][L2][P] = \frac{[MLT^{-2}]}{[L^2]}

Using the laws of indices, we subtract the exponent of the denominator from the numerator:

[P]=M×L(1−2)×T−2[P] = M \times L^{(1-2)} \times T^{-2}
[P]=ML−1T−2[P] = ML^{-1}T^{-2}
4. Identify the answer

Comparing our result with the given options: A. ML−1T−2ML^{-1}T^{-2} B. MLT−2MLT^{-2} C. ML2T3ML^2T^3 D. ML−3ML^{-3}

The correct dimension is ML−1T−2ML^{-1}T^{-2}, which matches option A.

5. Tip

Exam Strategy: If you ever forget the dimension of pressure, remember that pressure is the same as Stress (Force/Area) or Young's Modulus. Also, remember that MLT−2MLT^{-2} is the standard dimension for Force. Whenever you see "per unit area," simply divide the Force dimension by L2L^2.

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