1. Write dimensions of each quantity

We need dimensions of $Z=\frac{S^2\sigma}{\kappa}$ where:

• $S$ = e.m.f. per unit temperature difference
• $\sigma$ = electrical conductivity
• $\kappa$ = thermal conductivity

2. Dimension of e.m.f.

Potential difference (e.m.f.) is $V=\frac{\text{work}}{\text{charge}}$

Now,

• work $=[ML^2T^{-2}]$
• charge $=[IT]$

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

Since $S$ is e.m.f. per unit temperature difference, $[S]=[ML^2T^{-3}I^{-1}K^{-1}]$

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

3. Dimension of electrical conductivity $\sigma$

Electrical conductivity is reciprocal of resistivity.

Using $R=\frac{V}{I}$ so $[R]=\frac{[ML^2T^{-3}I^{-1}]}{[I]}=[ML^2T^{-3}I^{-2}]$

From $\rho = \frac{RA}{l}$ we get $[\rho]=[R][L]=[ML^3T^{-3}I^{-2}]$

Hence conductivity $[\sigma]=[\rho]^{-1}=[M^{-1}L^{-3}T^3I^2]$

4. Dimension of thermal conductivity $\kappa$

From Fourier's law, $\frac{Q}{t}=\kappa A\frac{\Delta T}{l}$ more precisely, $\frac{Q}{t}=\kappa A\frac{d\theta}{dx}$

So, $[\kappa]=\frac{[Q/t]\,[l]}{[A]\,[K]}$

Now,

• $[Q/t] =$ power $= [ML^2T^{-3}]$
• $[l]=[L]$
• $[A]=[L^2]$

Thus, $[\kappa]=\frac{[ML^2T^{-3}]\,[L]}{[L^2]\,[K]}=[MLT^{-3}K^{-1}]$

5. Now find dimension of $Z$

$[Z]=\frac{[S^2][\sigma]}{[\kappa]}$

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

First multiply numerator: $[S^2\sigma]=[M^{2-1}L^{4-3}T^{-6+3}I^{-2+2}K^{-2}]$ $=[MLT^{-3}K^{-2}]$

Now divide by $[\kappa]=[MLT^{-3}K^{-1}]$: $[Z]=[M^{1-1}L^{1-1}T^{-3-(-3)}I^0K^{-2-(-1)}]$ $[Z]=[M^0L^0T^0I^0K^{-1}]$

6. Match with options

This corresponds to: $[M^0L^0T^0I^0K^{-1}]$ which is Option B.

7. Comparison with stored answer

Stored correct answer: B

My derived answer: B

So, the derived answer agrees with the stored correct answer.