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, $\sigma$ the electrical conductivity and $k$ 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 $Z = \frac{S^{2}\sigma}{k}$ is:

Step 1: Find the dimensional formula of e.m.f. and $S$.

The electromotive force (e.m.f.) is defined as work done per unit charge.

$$[\text{e.m.f.}] = \frac{[\text{Work}]}{[\text{Charge}]}$$ $$[\text{e.m.f.}] = \frac{[ML^{2}T^{-2}]}{[IT]} = [ML^{2}T^{-3}I^{-1}]$$

Since $S$ is the e.m.f. produced per unit temperature difference:

$$[S] = \frac{[\text{e.m.f.}]}{[K]} = [ML^{2}T^{-3}I^{-1}K^{-1}]$$

Squaring this gives:

$$[S^{2}] = [M^{2}L^{4}T^{-6}I^{-2}K^{-2}]$$

Step 2: Find the dimensional formula of electrical conductivity $\sigma$.

Electrical conductivity is the reciprocal of resistivity $\rho$, where resistance $R = \frac{V}{I}$.

$$[R] = \frac{[ML^{2}T^{-3}I^{-1}]}{[I]} = [ML^{2}T^{-3}I^{-2}]$$ $$[\rho] = \frac{[R][A]}{[L]} = \frac{[ML^{2}T^{-3}I^{-2}][L^{2}]}{[L]} = [ML^{3}T^{-3}I^{-2}]$$ $$[\sigma] = \frac{1}{[\rho]} = [M^{-1}L^{-3}T^{3}I^{2}]$$

Step 3: Find the dimensional formula of thermal conductivity $k$.

From Fourier's law of heat conduction:

$$\frac{dQ}{dt} = kA\frac{dT}{dx}$$ $$[k] = \frac{[\text{Power}][L]}{[A][K]}$$ $$[k] = \frac{[ML^{2}T^{-3}][L]}{[L^{2}][K]} = [MLT^{-3}K^{-1}]$$

Step 4: Calculate the dimensional formula of $Z = \frac{S^{2}\sigma}{k}$.

Substitute the dimensions of $S^{2}$, $\sigma$, and $k$ into the expression for $Z$.

$$[Z] = \frac{[M^{2}L^{4}T^{-6}I^{-2}K^{-2}][M^{-1}L^{-3}T^{3}I^{2}]}{[MLT^{-3}K^{-1}]}$$

Combine the exponents for each base dimension:

$$[Z] = [M^{2-1-1}L^{4-3-1}T^{-6+3+3}I^{-2+2}K^{-2+1}]$$ $$[Z] = [M^{0}L^{0}T^{0}I^{0}K^{-1}]$$

Step 5: Verify using the Wiedemann-Franz Law.

The Wiedemann-Franz law relates thermal conductivity $k$ and electrical conductivity $\sigma$ at temperature $T$ via the Lorenz number $L$:

$$\frac{k}{\sigma T} = L$$

The Lorenz number $L$ has the same dimensions as $S^{2}$.

$$[L] = [S^{2}]$$

Substitute this relation back into the expression for $Z$:

$$Z = \frac{S^{2}\sigma}{k} = \frac{S^{2}}{LT}$$ $$[Z] = \frac{1}{[T]} = [K^{-1}]$$

This confirms the dimensional formula:

$$[Z] = [M^{0}L^{0}T^{0}I^{0}K^{-1}]$$

Answer:

$**[M^{0}L^{0}T^{0}I^{0}K^{-1}]**$