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JEE Advanced 2005 · Units and Measurements

Which of the following set have different dimensions?

Answer: D. Dipole moment, Electric flux, Electric field

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Written solution

Step 1: Analyze Set (1) - Pressure, Young's modulus, Stress.

Pressure and stress are defined as force per unit area, and Young's modulus is stress divided by the dimensionless quantity strain.

[P]=[σ]=[F][A]=M L T−2L2=M L−1T−2[P] = [\sigma] = \frac{[F]}{[A]} = \frac{\text{M L T}^{-2}}{\text{L}^2} = \text{M L}^{-1}\text{T}^{-2} [Y]=[σ][strain]=M L−1T−2[Y] = \frac{[\sigma]}{[\text{strain}]} = \text{M L}^{-1}\text{T}^{-2}

All three quantities share the same dimensional formula.

Step 2: Analyze Set (2) - EMF, Potential difference, Electric potential.

These quantities are defined as work done per unit charge.

[ε]=[V]=[ϕ]=[W][q][\varepsilon] = [V] = [\phi] = \frac{[W]}{[q]} [ε]=M L2T−2A T=M L2T−3A−1[\varepsilon] = \frac{\text{M L}^2\text{T}^{-2}}{\text{A T}} = \text{M L}^2\text{T}^{-3}\text{A}^{-1}

All three quantities share the same dimensional formula.

Step 3: Analyze Set (3) - Heat, Work done, Energy.

Heat and work are forms of energy, which is defined as force multiplied by displacement.

[Q]=[W]=[E]=[F][L][Q] = [W] = [E] = [F][L] [Q]=(M L T−2)(L)=M L2T−2[Q] = (\text{M L T}^{-2})(\text{L}) = \text{M L}^2\text{T}^{-2}

All three quantities share the same dimensional formula.

Step 4: Analyze Set (4) - Dipole moment, Electric flux, Electric field.

Calculate the dimensional formula for each quantity individually.

[p]=[q][L]=M0L T A[p] = [q][L] = \text{M}^0\text{L T A} [Efield]=[F][q]=M L T−2A T=M L T−3A−1[E_{\text{field}}] = \frac{[F]}{[q]} = \frac{\text{M L T}^{-2}}{\text{A T}} = \text{M L T}^{-3}\text{A}^{-1} [Φ]=[Efield][L2]=M L3T−3A−1[\Phi] = [E_{\text{field}}][L^2] = \text{M L}^3\text{T}^{-3}\text{A}^{-1}

Since the dimensional formulas do not match, the quantities in this set have different dimensions.

Answer: (4) Dipole moment, Electric flux, Electric field

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