Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter $1\ \text{mm}$. Conductor $B$ is a hollow tube of outer diameter $2\ \text{mm}$ and inner diameter $1\ \text{mm}$. Find the ratio of resistance $R_A$ to $R_B$.

> [!TLDR] > Resistance depends on cross-sectional area for a fixed material and length. Conductor A has area $\pi (0.5)^2$, conductor B has area $\pi (1^2 - 0.5^2)$. Their ratio $R_A/R_B = 3$. The key idea here is that resistance $R$ is given by $R = \rho \frac{L}{A}$, where $\rho$ is resistivity (same material, so same $\rho$), $L$ is length (same for both), and $A$ is the cross-sectional area. So the ratio of resistances is simply the inverse ratio of areas: $R_A / R_B = A_B / A_A$. Let’s work through it step by step. 1. **Find the area of conductor A** — a solid wire of diameter $1\ \text{mm}$. Radius $r_A = 0.5\ \text{mm}$. Area $A_A = \pi r_A^2 = \pi (0.5)^2 = 0.25\pi\ \text{mm}^2$. 2. **Find the area of conductor B** — a hollow tube with outer diameter $2\ \text{mm}$ and inner diameter $1\ \text{mm}$. Outer radius $R_o = 1\ \text{mm}$, inner radius $R_i = 0.5\ \text{mm}$. The cross-sectional area is the area of the outer circle minus the area of the hollow part: $A_B = \pi R_o^2 - \pi R_i^2 = \pi (1^2 - 0.5^2) = \pi (1 - 0.25) = 0.75\pi\ \text{mm}^2$. 3. **Take the ratio of resistances**. Since $R \propto 1/A$, \[ \frac{R_A}{R_B} = \frac{A_B}{A_A} = \frac{0.75\pi}{0.25\pi} = \frac{0.75}{0.25} = 3. \] …

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter

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Concept understanding — Temperature Dependence of Resistance

Temperature Dependence of Resistance

Imagine you're trying to walk through a crowded market. When the market is cool and calm, people move slowly and you can weave through easily. Now imagine the same market on a hot, chaotic day — everyone is jostling, moving faster, bumping into each other. Getting from one end to the other becomes much harder.

That's exactly what happens inside a metal wire when you heat it up.

The Intuition

In a metal, electric current is carried by free electrons drifting through a fixed lattice of positive ions. At room temperature, these ions are vibrating slightly around their positions. When you heat the metal, the ions vibrate more vigorously — they shake faster and with larger amplitude.

Think of the vibrating ions as a row of swinging doors. At low temperature, the doors barely move, so electrons slip through easily. At high temperature, the doors swing wildly, and electrons get knocked off course constantly. Each collision with a vibrating ion scatters the electron, making it harder for the current to flow.

The result: resistance increases as temperature increases — for most conductors.

The Precise Statement

For a metallic conductor over a moderate temperature range (not too close to absolute zero), the resistance changes linearly with temperature:

R(T) = R_0 [1 + \alpha (T - T_0)]

Where:

$R(T)$ is the resistance at temperature $T$
$R_0$ is the resistance at a reference temperature $T_0$ (often $0^\circ$ C or $20^\circ$ C)
$\alpha$ is the temperature coefficient of resistance (units: per °C or per K)

R = R_0 (1 + \alpha \Delta T)

The coefficient $\alpha$ tells you how sensitive the material is to temperature changes. For copper, $\alpha \approx 0.0039 /^\circ\text{C}$ — meaning for every 1°C rise, resistance increases by about 0.39%.

What About Other Materials?

Not everything behaves like metals.

Semiconductors (like silicon, germanium) do the opposite: their resistance decreases sharply as temperature rises. Why? Because heating frees more electrons from their bonds, creating many more charge carriers. Even though the lattice vibrates more, the huge increase in available carriers overwhelms that effect, so resistance drops.

Insulators also show decreasing resistance with temperature, but the effect is much smaller than in semiconductors.

Alloys like constantan (copper-nickel) have a very small $\alpha$ — their resistance barely changes with temperature. This is useful for making precision resistors that stay stable.

Superconductors are a special case: below a critical temperature, resistance drops to exactly zero. …

Why this formula?

Temperature Dependence of Resistance — Why the Formula Holds

Let’s build this from the ground up. The key formula you’ll see in exams is:

R_T = R_0 (1 + \alpha T)

But why does resistance change with temperature? It’s not magic — it’s about what happens inside the wire.

1. What determines resistance?

Resistance $R$ of a conductor depends on three things:

Length $L$ (longer → more resistance)
Cross-sectional area $A$ (thicker → less resistance)
Resistivity $\rho$ — a material property

The formula is:

R = \rho \frac{L}{A}

When temperature changes, $L$ and $A$ change very slightly (thermal expansion), but the big effect is on $\rho$ .

2. Why does resistivity change with temperature?

Resistivity $\rho$ depends on how easily electrons can move through the material.

In metals: Atoms vibrate more as temperature rises. These vibrations scatter electrons, making it harder for them to flow. So $\rho$ increases.
In semiconductors: More electrons get enough energy to jump into the conduction band. So $\rho$ decreases.

For most metals (and many conductors), the change is linear over a moderate temperature range.

3. Deriving the linear formula

Let $\rho_0$ be resistivity at a reference temperature $T_0$ (often $0^\circ$ C or $20^\circ$ C).

For a small change $\Delta T = T - T_0$ , the change in resistivity is proportional to $\Delta T$ and to $\rho_0$ :

\Delta \rho \propto \rho_0 \Delta T

Introduce the temperature coefficient of resistivity $\alpha$ :

\Delta \rho = \alpha \, \rho_0 \, \Delta T

So the new resistivity is:

\rho = \rho_0 + \Delta \rho = \rho_0 (1 + \alpha \Delta T)

Now, since $R = \rho \frac{L}{A}$ , and $L$ and $A$ change negligibly (for small $\Delta T$ ), we get:

R = \rho \frac{L}{A} = \rho_0 (1 + \alpha \Delta T) \frac{L}{A} = R_0 (1 + \alpha \Delta T)

That’s the formula:

\boxed{R_T = R_0 (1 + \alpha \Delta T)}

Where:

$R_T$ = resistance at temperature $T$
$R_0$ = resistance at reference temperature $T_0$
$\alpha$ = temperature coefficient of resistance (unit: $^\circ$ C $^{-1}$ or K $^{-1}$ )
$\Delta T = T - T_0$

--- …

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Resistance depends on cross-sectional area for a fixed material and length. Conductor A has area $\pi (0.5)^2$ , conductor B has area $\pi (1^2 - 0.5^2)$ . Their ratio $R_A/R_B = 3$ .

The key idea here is that resistance $R$ is given by $R = \rho \frac{L}{A}$ , where $\rho$ is resistivity (same material, so same $\rho$ ), $L$ is length (same for both), and $A$ is the cross-sectional area. So the ratio of resistances is simply the inverse ratio of areas: $R_A / R_B = A_B / A_A$ .

Let’s work through it step by step.

Find the area of conductor A — a solid wire of diameter $1\ \text{mm}$ .

Radius $r_A = 0.5\ \text{mm}$ .

Area $A_A = \pi r_A^2 = \pi (0.5)^2 = 0.25\pi\ \text{mm}^2$ .
Find the area of conductor B — a hollow tube with outer diameter $2\ \text{mm}$ and inner diameter $1\ \text{mm}$ .

Outer radius $R_o = 1\ \text{mm}$ , inner radius $R_i = 0.5\ \text{mm}$ .

The cross-sectional area is the area of the outer circle minus the area of the hollow part:

$A_B = \pi R_o^2 - \pi R_i^2 = \pi (1^2 - 0.5^2) = \pi (1 - 0.25) = 0.75\pi\ \text{mm}^2$ .
Take the ratio of resistances.

Since $R \propto 1/A$ ,

$\frac{R_A}{R_B} = \frac{A_B}{A_A} = \frac{0.75\pi}{0.25\pi} = \frac{0.75}{0.25} = 3.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Which of the options given below has resistivity that decreases with increase in temperature?$
(a) Metal
(b) Alloy
(c) Semiconductor
(d) Insulator
›Reveal solutionSolution

Metals and alloys have a positive temperature coefficient of resistivity (resistivity rises with temperature), while semiconductors have a negative temperature coefficient.

In a semiconductor, rising temperature releases more charge carriers across the band gap (increasing carrier density n much faster than the mobility drops), so overall resistivity falls s …

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GUJCET 2024Set 131 markMCQ
$2.1\,\Omega$
›Reveal solutionSolution

$\alpha=\dfrac{R_2-R_1}{R_1(T_2-T_1)}=\dfrac{0.6}{2.1\times72.5}\approx3.9\times10^{-3}\ ^\circ\text{C}^{-1}$ .

Concept. Temperature coefficient of resistance $\alpha=\dfrac{\Delta R}{R_1\,\Delta T}$ (units are per degree Celsius).

Steps. $\Delta R = 2.7-2.1 = 0.6\,\Omega$ , $\Delta T = 100-27.5 = 72.5\,^\circ$ C: …
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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. Resistivity of which of the following substance decrease on increasing the temperature?$
(a) Copper
(b) Silicon
(c) Aluminium
(d) Nichrome
›Reveal solutionSolution

Metals (Cu, Al, Nichrome) have resistivity that increases with temperature, while semiconductors show the opposite behaviour.

In metals, increasing temperature increases lattice vibrations, scattering more electrons and raising resistivity. In semiconductors like silicon, increasing temperature releases more charge carriers across the band ga …

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GUJCET 2022Set 171 markMCQ
$27\degree$
›Reveal solutionSolution

From $R=R_0[1+\alpha(T-T_0)]$ : $T-27=\dfrac{0.37}{1.35\times10^{-4}}\approx2741$ , so $T\approx2767^\circ$ C.

Concept: $R=R_0[1+\alpha(T-T_0)]$ .

$\frac{137}{100}=1+\alpha(T-27)\Rightarrow 0.37=(1.35\times10^{-4})(T-27)$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. For metals, the value of the temperature coefficient of resistivity (α) is ______.$
(a) zero
(b) positive
(c) negative
(d) infinite
›Reveal solutionSolution

Metals get more resistive as they heat up, so their temperature coefficient of resistivity is positive.

For metals, resistivity follows $\rho_T = \rho_0[1 + \alpha(T - T_0)]$ . As temperature rises, increased thermal vibration of the lattice ions scatters conduction electrons more frequently, reducing the relaxation time and increasing …

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GUJCET 2020Set 071 markMCQ
$5\,\Omega$
›Reveal solutionSolution

Using the platinum-resistance linear scale, $t=\dfrac{R_t-R_0}{R_{100}-R_0}\times100 = 345.65\,^\circ C$ .

Concept. For a platinum resistance thermometer,
$t=\frac{R_t-R_0}{R_{100}-R_0}\times100\,^\circ C,$
with $R_0$ at ice point and $R_{100}$ at steam point.

Compute. …
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Q.Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter $1\ \text{mm}$ . Conductor $B$ is a hollow tube of outer diameter $2\ \text{mm}$ and inner diameter $1\ \text{mm}$ . Find the ratio of resistance $R_A$ to $R_B$ .

Concept understanding — Temperature Dependence of Resistance

Temperature Dependence of Resistance

The Intuition

The Precise Statement

What About Other Materials?

Why this formula?

Temperature Dependence of Resistance — Why the Formula Holds

1. What determines resistance?

2. Why does resistivity change with temperature?

3. Deriving the linear formula

Method: Finding Resistance Ratios from Cross-Sectional Geometry

Steps

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