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PhyMedAdv 2025 • Paper 2

Dimensional Formula of Thermoelectric Figure of Merit Quantity

A temperature difference can generate e.m.f. in some materials. Let SS be the e.m.f. produced per unit temperature difference between the ends of a wire, σ\sigma the electrical conductivity and κ\kappa the thermal conductivity of the material of the wire. Taking M,L,T,IM, L, T, I and KK as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity Z=S2σκZ = \frac{S^2 \sigma}{\kappa} is:

Options

A

[M0L0T0I0K0][M^0 L^0 T^0 I^0 K^0]

B

[M0L0T0I0K−1][M^0 L^0 T^0 I^0 K^{-1}]

Correct
C

[M1L2T−2I−1K−1][M^1 L^2 T^{-2} I^{-1} K^{-1}]

D

[M1L2T−4I−1K−1][M^1 L^2 T^{-4} I^{-1} K^{-1}]

Topics & Concepts

Units and MeasurementsOhm's LawHeat Transfer

Step-by-Step Solution

To find the dimensional formula of the thermoelectric figure of merit quantity Z=S2σκZ = \frac{S^2 \sigma}{\kappa}, we first determine the dimensions of each constituent physical quantity: SS, σ\sigma, and κ\kappa.

  1. Dimensions of Seebeck Coefficient (SS): The Seebeck coefficient SS is defined as the electromotive force (e.m.f.) produced per unit temperature difference: S=VΔTS = \frac{V}{\Delta T} The dimension of electric potential (or e.m.f.) VV is given by: [V]=[Work][Charge]=[ML2T−2][IT]=[ML2T−3I−1][V] = \frac{[\text{Work}]}{[\text{Charge}]} = \frac{[M L^2 T^{-2}]}{[I T]} = [M L^2 T^{-3} I^{-1}] Therefore, the dimension of SS is: [S]=[ML2T−3I−1][K]=[ML2T−3I−1K−1][S] = \frac{[M L^2 T^{-3} I^{-1}]}{[K]} = [M L^2 T^{-3} I^{-1} K^{-1}]

  2. Dimensions of Electrical Conductivity (σ\sigma): Electrical conductivity is the reciprocal of resistivity (ρ\rho): σ=lRA\sigma = \frac{l}{R A} The dimension of electrical resistance RR is: [R]=[V][I]=[ML2T−3I−1][I]=[ML2T−3I−2][R] = \frac{[V]}{[I]} = \frac{[M L^2 T^{-3} I^{-1}]}{[I]} = [M L^2 T^{-3} I^{-2}] Therefore, the dimension of electrical conductivity σ\sigma is: [σ]=[L][ML2T−3I−2]⋅[L2]=[M−1L−3T3I2][\sigma] = \frac{[L]}{[M L^2 T^{-3} I^{-2}] \cdot [L^2]} = [M^{-1} L^{-3} T^3 I^2]

  3. Dimensions of Thermal Conductivity (κ\kappa): Thermal conductivity is defined through the heat transfer equation: dQdt=κAΔTl\frac{dQ}{dt} = \kappa A \frac{\Delta T}{l} Here, dQdt\frac{dQ}{dt} represents power, with dimensions [ML2T−3][M L^2 T^{-3}]. Thus, the dimension of thermal conductivity κ\kappa is: [κ]=[dQdt][L][A][ΔT]=[ML2T−3][L][L2][K]=[MLT−3K−1][\kappa] = \frac{\left[\frac{dQ}{dt}\right] [L]}{[A] [\Delta T]} = \frac{[M L^2 T^{-3}] [L]}{[L^2] [K]} = [M L T^{-3} K^{-1}]

  4. Dimensional Formula of ZZ: Now, substituting the dimensions into Z=S2σκZ = \frac{S^2 \sigma}{\kappa}: [S2]=([ML2T−3I−1K−1])2=[M2L4T−6I−2K−2][S^2] = \left([M L^2 T^{-3} I^{-1} K^{-1}]\right)^2 = [M^2 L^4 T^{-6} I^{-2} K^{-2}]

    Combining [S2][S^2] and [σ][\sigma]: [S2σ]=[M2L4T−6I−2K−2]×[M−1L−3T3I2]=[M1L1T−3I0K−2][S^2 \sigma] = [M^2 L^4 T^{-6} I^{-2} K^{-2}] \times [M^{-1} L^{-3} T^3 I^2] = [M^1 L^1 T^{-3} I^0 K^{-2}]

    Finally, dividing by [κ][\kappa]: [Z]=[M1L1T−3K−2][M1L1T−3K−1]=[M0L0T0I0K−1][Z] = \frac{[M^1 L^1 T^{-3} K^{-2}]}{[M^1 L^1 T^{-3} K^{-1}]} = [M^0 L^0 T^0 I^0 K^{-1}]

Thus, the dimensional formula of the quantity ZZ is [M0L0T0I0K−1][M^0 L^0 T^0 I^0 K^{-1}], which corresponds to Option B.

Dimensional Formula of Thermoelectric Figure of Merit Quantity | Physics PYQ Solution - JEE Challenger