Question 1

We first find the dimensions of each quantity:

1. e.m.f. ($V$): $\text{work} / \text{charge} = [ML^2T^{-2}] / [IT] = [ML^2T^{-3}I^{-1}]$.
2. $S = \text{e.m.f.} / \text{temperature} = [ML^2T^{-3}I^{-1}K^{-1}]$.
3. Electrical conductivity $\sigma = \frac{1}{\rho} = \frac{l}{RA} = \frac{l}{(V/I)A} = \frac{[L][I]}{[ML^2T^{-3}I^{-1}][L^2]} = [M^{-1}L^{-3}T^3I^2]$.
4. Thermal conductivity $\kappa$ is found from $\frac{dQ}{dt} = \kappa A \frac{\Delta T}{l} \Rightarrow \kappa = \frac{[ML^2T^{-2}/T] [L]}{[L^2][K]} = [MLT^{-3}K^{-1}]$.

Now, calculate dimensions of $Z = \frac{S^2 \sigma}{\kappa}$: $[Z] = \frac{([ML^2T^{-3}I^{-1}K^{-1}])^2 \cdot [M^{-1}L^{-3}T^3I^2]}{[MLT^{-3}K^{-1}]}$ $[Z] = \frac{[M^2L^4T^{-6}I^{-2}K^{-2}] \cdot [M^{-1}L^{-3}T^3I^2]}{[MLT^{-3}K^{-1}]}$ $[Z] = \frac{[M^1L^1T^{-3}K^{-2}]}{[MLT^{-3}K^{-1}]} = [K^{-1}]$ Thus, the dimensional formula is $[M^0 L^0 T^0 I^0 K^{-1}]$.