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JEE Advanced 2025 · Paper 2, Physics question 1

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 = \dfrac{S^2\sigma}{\kappa} is:

Choose one option
Answer and worked solution

Official answer: B

Dimensions of each quantity.

  • SS = volts per kelvin, and [V]=ML2T−3I−1[V]=ML^2T^{-3}I^{-1}, so [S2]=M2L4T−6I−2K−2[S^2]=M^2L^4T^{-6}I^{-2}K^{-2}.
  • [σ]=M−1L−3T3I2[\sigma]=M^{-1}L^{-3}T^3I^2 (siemens per metre).
  • [κ]=MLT−3K−1[\kappa]=MLT^{-3}K^{-1} (W m⁻¹ K⁻¹).

Combine.
[S2σκ]=MLT−3K−2MLT−3K−1=K−1.\left[\frac{S^2\sigma}{\kappa}\right]=\frac{MLT^{-3}K^{-2}}{MLT^{-3}K^{-1}}=K^{-1}.

That is [M0L0T0I0K−1][M^0L^0T^0I^0K^{-1}]. Multiplied by temperature, ZZ gives the dimensionless thermoelectric figure of merit ZTZT.

Topics:Dimensional analysis
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