JEE Advance - Physics (2025 - Paper 2 Online - No. 1)
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 of the wire. Taking M, L, T, I and K as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $Z = \frac{S^2 \sigma}{\kappa}$ is :
[M0L0T0I0K0]
[M0L0T0I0K−1]
[M1L2T−2I−1K−1]
[M1L2T−4I−1K−1]
Explanation
S is emf per unit temperature ⇒
$$[S]=\frac{[V]}{[K]}=\frac{M\,L^2\,T^{-3}\,I^{-1}}{K} =M\,L^2\,T^{-3}\,I^{-1}\,K^{-1}.$$
σ is electrical conductivity ⇒
$$[σ]=\frac{[J]}{[E]} =\frac{I\,L^{-2}}{M\,L\,T^{-3}\,I^{-1}} =M^{-1}\,L^{-3}\,T^3\,I^2.$$
κ is thermal conductivity ⇒
$$[κ]=\frac{\dot Q/(A)}{∇T} =\frac{M\,L^2\,T^{-3}/L^2}{K\,L^{-1}} =M\,L\,T^{-3}\,K^{-1}.$$
Now
$$[Z] =[S]^2\,[σ]\,[κ]^{-1} =(M\,L^2\,T^{-3}\,I^{-1}\,K^{-1})^2 \times M^{-1}\,L^{-3}\,T^3\,I^2 \times (M\,L\,T^{-3}\,K^{-1})^{-1}$$
Simplifying exponents gives
$$[Z]=M^0\,L^0\,T^0\,I^0\,K^{-1}.$$
Answer: Option B.
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