$\text{A temperature difference can generate e.m.f. in some materials. Let}S\text{be the e.m.f. produced per unit temperature}$

$\text{difference between the ends of a wire,}\sigma \text{the electrical conductivity and}\kappa \text{the thermal conductivity of the material}$

$\text{of the wire. Taking}\mathrm{ M},L,T,I\text{ and}\mathrm{ K}\text{ as dimensions of mass, length, time, current and temperature, respectively,}$

$\text{the dimensional formula of the quantity}Z=\frac{S^2\sigma }{\kappa }\text{is:}$

The correct answer is:
$\left[M^0{L}^0{T}^0{I}^0{K}^{-1}\right]$ (or $\left[M^0{L}^0{T}^0{\theta }^0{K}^{-1}\right]$ , where $\theta$ or $I$ represents electrical current).

Step-by-Step Derivation:

We are given the quantity:
$Z=\frac{S^2\sigma }{\kappa }$
We need to determine the dimensional formula of each of the physical variables: $S$, $\sigma$, and $\kappa$.

1. Dimensional Formula of $S$ (e.m.f. per unit temperature difference):
The electromotive force (e.m.f.) $V$ is work done per unit charge:
$\left[V\right]=\frac{\left[\text{Work}\right]}{\left[\text{Charge}\right]}=\frac{M{L}^2{T}^{-2}}{IT}=M{L}^2{T}^{-3}{I}^{-1}$
Since $S$ is e.m.f. per unit temperature difference:
$\left[S\right]=\frac{\left[V\right]}{\left[\text{Temperature}\right]}=\frac{M{L}^2{T}^{-3}{I}^{-1}}{K}=M{L}^2{T}^{-3}{I}^{-1}{K}^{-1}$

2. Dimensional Formula of $\sigma$ (electrical conductivity):
Electrical conductivity is the reciprocal of resistivity ($\rho$):
$\sigma =\frac{1}{\rho }=\frac{l}{RA}$
where $R$ is resistance, $l$ is length, and $A$ is area. Resistance is given by Ohm's Law:
$\left[R\right]=\frac{\left[V\right]}{\left[I\right]}=\frac{M{L}^2{T}^{-3}{I}^{-1}}{I}=M{L}^2{T}^{-3}{I}^{-2}$
Thus, the dimensions of electrical conductivity $\sigma$ are:
$\left[\sigma \right]=\frac{L}{\left(M{L}^2{T}^{-3}{I}^{-2}\right)\left(L^2\right)}=M^{-1}{L}^{-3}{T}^3{I}^2$

3. Dimensional Formula of $\kappa$ (thermal conductivity):
Thermal conductivity is defined by the rate of heat flow equation:
$\frac{Q}{t}=\frac{\kappa A\Delta T}{d}$
where $Q$ is thermal energy, $t$ is time, $A$ is area, $\Delta T$ is temperature difference, and $d$ is thickness. Rearranging for $\kappa$:
$\left[\kappa \right]=\frac{\left[Q\right]\cdot \left[d\right]}{\left[t\right]\cdot \left[A\right]\cdot \left[\Delta T\right]}=\frac{\left(M{L}^2{T}^{-2}\right)\cdot L}{T\cdot L^2\cdot K}=ML{T}^{-3}{K}^{-1}$

4. Dimensional Formula of $Z$ :
Substituting these values back into the expression for $Z$:
$\left[Z\right]=\frac{{\left[S\right]}^2\left[\sigma \right]}{\left[\kappa \right]}$
Let us first evaluate the numerator:
${\left[S\right]}^2=M^2{L}^4{T}^{-6}{I}^{-2}{K}^{-2}$
Multiplying by $\left[\sigma \right]$:
${\left[S\right]}^2\left[\sigma \right]=\left(M^2{L}^4{T}^{-6}{I}^{-2}{K}^{-2}\right)\cdot \left(M^{-1}{L}^{-3}{T}^3{I}^2\right)$
${\left[S\right]}^2\left[\sigma \right]=M^{2-1}{L}^{4-3}{T}^{-6+3}{I}^{-2+2}{K}^{-2}=ML{T}^{-3}{I}^0{K}^{-2}$
Now, dividing by the denominator $\left[\kappa \right]$:
$\left[Z\right]=\frac{ML{T}^{-3}{K}^{-2}}{ML{T}^{-3}{K}^{-1}}=M^0{L}^0{T}^0{I}^0{K}^{-1}$