A temperature difference can generate e.m.f. in some materials. Let S be the e.m.f. produced per uni

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Question

A temperature difference can generate e.m.f. in some materials. Let S be the e.m.f. produced per unit temperature difference between the ends of a wire, σ the electrical conductivity and κ the thermal conductivity of the material. Taking M, L, T, I and K as dimensions of mass, length, time, current and temperature respectively, the dimensional formula of the quantity Z = S²σ/κ is:

Accepted Answer

Correct Answer: B — [M⁰ L⁰ T⁰ I⁰ K⁻¹]Step-by-step Solution:We need to find [Z] = [S²σ/κ].Dimensions of S (Seebeck coefficient = EMF per unit temperature):[S] = [EMF]/[T] = [Voltage]/[K] = [ML²T⁻³I⁻¹]/[K] = [ML²T⁻³I⁻¹K⁻¹][S²] = [M²L⁴T⁻⁶I⁻²K⁻²]Dimensions of σ (electrical conductivity = 1/resistivity):[Resistivity ρ] = [Ω·m] = [ML³T⁻³I⁻²][σ] = 1/[ρ] = [M⁻¹L⁻³T³I²]Dimensions of κ (thermal conductivity):κ is defined by Fourier's law: Q/t = κ·A·(ΔT/L)[κ] = [Q·L/(t·A·ΔT)] = [ML²T⁻²·L/(T·L²·K)] = [ML T⁻³ K⁻¹]Computing [Z] = [S²σ/κ]:[Z] = [M²L⁴T⁻⁶I⁻²K⁻²] · [M⁻¹L⁻³T³I²] / [ML T⁻³ K⁻¹]Numerator: [M²·M⁻¹ · L⁴·L⁻³ · T⁻⁶·T³ · I⁻²·I² · K⁻²] = [M¹ L¹ T⁻³ I⁰ K⁻²]Dividing by [κ] = [ML T⁻³ K⁻¹]:[Z] = [M¹L¹T⁻³K⁻²] / [MLT⁻³K⁻¹] = [M⁰ L⁰ T⁰ K⁻¹] = [M⁰ L⁰ T⁰ I⁰ K⁻¹]Why other options are incorrect:Option A [M⁰L⁰T⁰I⁰K⁰] (dimensionless): One would get this if the K dimensions canceled completely. But [S²] contributes K⁻² and [κ] contributes K⁻¹, so the net K-dimension is K⁻²/K⁻¹ = K⁻¹, not K⁰.Option C [M⁻¹L⁻²T²IK⁻¹]: This would arise from errors in computing [σ] or [S]. For example, using [σ] = [M⁻¹L⁻²T²I²] (missing one factor of L in resistivity) would shift the L and M dimensions.Option D [M⁻¹L⁻²T⁴IK⁻¹]: The extra T factors (T⁴ instead of T⁰) suggest an error in combining the T-dimensions of S², σ, and κ. For example, using [S] = [ML²T⁻²I⁻¹K⁻¹] (energy instead of voltage) gives wrong T-dimension.Key Concepts:Seebeck coefficient S: S = EMF/ΔT = Volts/Kelvin = [ML²T⁻³I⁻¹K⁻¹]. This is a fundamental thermoelectric quantity relating to the Seebeck effect (used in thermocouples).Electrical conductivity σ: σ = 1/ρ where ρ is resistivity. [ρ] = [Ω·m] = [ML³T⁻³I⁻²], so [σ] = [M⁻¹L⁻³T³I²].Thermal conductivity κ: From Fourier's law Q/t = κA(dT/dx): [κ] = [W/(m·K)] = [MLT⁻³K⁻¹].Figure of merit ZT in thermoelectrics: The dimensionless quantity ZT = S²σT/κ is the thermoelectric figure of merit. Since ZT is dimensionless and T has dimension [K], this confirms that Z = S²σ/κ has dimension [K⁻¹], exactly as computed above. High ZT (> 1) indicates good thermoelectric performance.

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