A temperature difference can generate e.m.f. in some materials. Let S be the e.m.f. produced per uni

Q: Why is the dimension of Voltage $ML^2T^{-3}I^{-1}$?

Voltage is Work done per unit charge ($V = W/q$). Work has dimensions of Energy ($ML^2T^{-2}$), and charge is Current $\times$ Time ($IT$). Dividing them gives $ML^2T^{-3}I^{-1}$.

Q: How do I remember the formula for thermal conductivity $\kappa$?

Think of Fourier's Law of heat conduction: The rate of heat flow ($P = Q/t$) is proportional to the Area ($A$) and the temperature gradient ($\Delta T/L$). So, $P = \kappa A \Delta T / L$. Rearranging this gives $\kappa = P L / (A \Delta T)$.

Q: What is the physical meaning of the quantity $Z$?

$Z$ is related to the 'Thermoelectric Figure of Merit', usually denoted as $ZT$ (where $T$ is absolute temperature). Since $ZT$ is a dimensionless number used to measure the efficiency of thermoelectric materials, $Z$ itself must have the dimension of inverse temperature ($K^{-1}$).

Q: Can I solve this without deriving every single dimension from scratch?

If you remember that $Z \times T$ is a dimensionless constant (Figure of Merit) from advanced thermodynamics, you can instantly guess the answer is $K^{-1}$. However, for JEE, deriving it ensures you don't rely on memorizing obscure constants.

Q: I got confused with the powers of $T$ and $K$. How to avoid this?

Always write $T$ for Time and $K$ (or $\theta$) for Temperature clearly. When substituting, do it step-by-step. First, find the numerator $S^2\sigma$, simplify it completely, and then divide by the denominator $\kappa$.

Question

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 $\kappa$ 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 $Z = \frac{S^2\sigma}{\kappa}$ is :-

Options:

(A) $[M^0L^0T^0I^0K^0]$

(B) $[M^0L^0T^0I^0K^{-1}]$

(C) $[M^1L^2T^{-2}I^{-1}K^{-1}]$

(D) $[M^1L^2T^{-4}I^{-1}K^{-1}]$

Sigma Percentile · Accepted Answer

### The Thermoelectric Phenomenon

Imagine a simple conducting wire. When you heat one end and keep the other end cold, something magical happens. The temperature difference causes electrons to migrate, generating an electromotive force (emf) across the wire. This is known as the **Seebeck effect**.

In this problem, we are exploring a specific quantity $Z$ that combines three fundamental properties of this wire: the Seebeck coefficient $S$, the electrical conductivity $\sigma$, and the thermal conductivity $\kappa$.

Our mission is to find the dimensional formula for $Z = \frac{S^2\sigma}{\kappa}$. To conquer this, we must break down each property into its fundamental dimensions: Mass ($M$), Length ($L$), Time ($T$), Current ($I$), and Temperature ($K$).

### Deconstructing the Dimensions

Let's tackle them one by one. First up is $S$, the emf produced per unit temperature difference.

Since emf is essentially voltage ($V$), and voltage is work done per unit charge ($W/q$), we can write:
$$S = \frac{V}{\Delta T} = \frac{W}{q \Delta T}$$

Substituting the basic dimensions, we get:
$$[S] = \frac{[ML^2T^{-2}]}{[IT][K]} = [M^1L^2T^{-3}I^{-1}K^{-1}]$$

Next, we have the electrical conductivity $\sigma$. We know it is the reciprocal of resistivity ($ho$). Using the classic resistance formula $R = \frac{ho L}{A}$, we can express conductivity as:
$$\sigma = \frac{1}{ho} = \frac{L}{RA} = \frac{I \cdot L}{V \cdot A}$$

Plugging in the dimensions (and using the dimension of voltage we just found), we get:
$$[\sigma] = \frac{[I][L]}{[ML^2T^{-3}I^{-1}][L^2]} = [M^{-1}L^{-3}T^3I^2]$$

Finally, let's find the dimension of thermal conductivity $\kappa$. Recall Fourier's law of heat conduction, where the rate of heat flow ($P = \frac{dQ}{dt}$) is given by:
$$P = \frac{\kappa A \Delta T}{L} \implies \kappa = \frac{P \cdot L}{A \cdot \Delta T}$$

Since power $P$ has dimensions of $[ML^2T^{-3}]$, we find:
$$[\kappa] = \frac{[ML^2T^{-3}][L]}{[L^2][K]} = [M^1L^1T^{-3}K^{-1}]$$

### The Grand Assembly

Now comes the most satisfying part. We substitute these three hard-earned dimensional formulas back into our original expression for $Z$:
$$[Z] = \frac{[S]^2[\sigma]}{[\kappa]}$$

First, let's square the dimension of $S$:
$$[S]^2 = [M^2L^4T^{-6}I^{-2}K^{-2}]$$

Now, multiply this by the dimension of $\sigma$ to get the numerator:
$$[S]^2[\sigma] = [M^2L^4T^{-6}I^{-2}K^{-2}] 	imes [M^{-1}L^{-3}T^3I^2]$$

Notice how beautifully the terms combine! The powers of $I$ cancel out completely ($I^{-2} \cdot I^2 = I^0$). The numerator simplifies to:
$$	ext{Numerator} = [M^1L^1T^{-3}K^{-2}]$$

### The Final Cancellation

Finally, we divide this numerator by the dimension of $\kappa$:
$$[Z] = \frac{[M^1L^1T^{-3}K^{-2}]}{[M^1L^1T^{-3}K^{-1}]}$$

Look at that! The $M$, $L$, and $T$ terms are identical in the numerator and denominator. They annihilate each other perfectly. We are left with:
$$[Z] = [M^0L^0T^0I^0K^{-1}]$$

**The final dimensional formula is simply $[K^{-1}]$.**

Interestingly, this isn't just a random math exercise. The quantity $Z$ is a crucial parameter in physics known as the **Thermoelectric Figure of Merit**. It is often multiplied by the absolute temperature $T$ to form the dimensionless quantity $ZT$, which determines how efficient a material is at converting heat into electricity!

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Visualized Solution

The Thermoelectric Setup

Target Quantity $Z$

Dimension of $S$

Dimension of $σ$

Dimension of $κ$

Substituting Dimensions

Final Calculation

The Sigma Insight: Dimensional Analysis

The Thermoelectric Phenomenon

Deconstructing the Dimensions

The Grand Assembly

The Final Cancellation

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Visualized Solution

The Thermoelectric Setup

Target Quantity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z

Dimension of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Dimension of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ

Dimension of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } κ