A temperature difference can generate e.m.f. | vedclass

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(A) The dimension of $e.m.f.$ is $[M L^2 T^{-3} I^{-1}]$. Since $S$ is $e.m.f.$ per unit temperature difference,$[S] = [M L^2 T^{-3} I^{-1} K^{-1}]$.E

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$\left[M^0 L^0 T^0 I^0 K^{-1}\right]$

Vedclass · Answer

(A) The dimension of $e.m.f.$ is $[M L^2 T^{-3} I^{-1}]$. Since $S$ is $e.m.f.$ per unit temperature difference,$[S] = [M L^2 T^{-3} I^{-1} K^{-1}]$.Electrical conductivity $\sigma$ is the reciprocal of resistivity $\rho$. Since $R = \rho \frac{l}{A}$,$\rho = \frac{R A}{l}$. The dimension of resistance $R$ is $[M L^2 T^{-3} I^{-2}]$. Thus,$[\rho] = [M L^3 T^{-3} I^{-2}]$ and $[\sigma] = [M^{-1} L^{-3} T^3 I^2]$.Thermal conductivity $\kappa$ is defined by $Q = \frac{\kappa A (T_2 - T_1) t}{d}$,so $[\kappa] = \frac{[Energy] [Length]}{[Area] [Temperature] [Time]} = [M L T^{-3} K^{-1}]$.Now,calculate the dimension of $Z = \frac{S^2 \sigma}{\kappa}$:$[Z] = \frac{[M L^2 T^{-3} I^{-1} K^{-1}]^2 [M^{-1} L^{-3} T^3 I^2]}{[M L T^{-3} K^{-1}]}$$[Z] = \frac{[M^2 L^4 T^{-6} I^{-2} K^{-2}] [M^{-1} L^{-3} T^3 I^2]}{[M L T^{-3} K^{-1}]}$$[Z] = \frac{[M L T^{-3} K^{-2}]}{[M L T^{-3} K^{-1}]} = [K^{-1}] = [M^0 L^0 T^0 I^0 K^{-1}]$.

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