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.

KEAM 2026Set eng-2026-04184 marksMCQ
$4\,\Omega$
›Reveal solutionSolution

Using $R_t=R_0(1+\alpha t)$ at the two temperatures and eliminating $R_0$ gives $\alpha=0.02\,{}^{\circ}C^{-1}$ .

With $R_t=R_0(1+\alpha t)$ :
$R_1=R_0(1+\alpha t_1),\qquad R_2=R_0(1+\alpha t_2)$
Eliminating $R_0$ :

$\alpha=\frac{R_2-R_1}{R_1 t_2-R_2 t_1}$ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$4 \times 10^{-3} K^{-1}$
›Reveal solutionSolution

Linear resistance–temperature law: $R=R_0(1+\alpha\Delta T)$ gives $12.4\ \Omega$ .

Compute. $\Delta T=80-20=60\,^\circ$ C. …
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KEAM 2026Set eng-2026-04224 marksMCQ
$Q. With increase in temperature, the resistivity of (A) both Cu and Ge increases (B) both nichrome and Ge decreases (C) nichrome decreases and that of Ge increases (D) both Cu and nichrome increases (E) Cu, nichrome and Ge increases$
›Reveal solutionSolution

Metals and alloys have positive temperature coefficients; semiconductors (Ge) have negative.

With rising temperature: Cu (a metal) and nichrome (an alloy) both show increasing resistivity, while germanium (a semiconductor) shows decreasing resistivity. …
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KEAM 2025Set eng-2025-04264 marksMCQ
$\Omega$
›Reveal solutionSolution

The temperature coefficient of resistance is $0.05\,/^{\circ}C$ .

Using $R=R_0(1+\alpha t)$ with $R_{30}=5\,\Omega$ and $R_{40}=6\,\Omega$ :
$\dfrac{R_{40}}{R_{30}}=\dfrac{1+40\alpha}{1+30\alpha}=\dfrac{6}{5}.$
Cross-multiplying: …
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KEAM 2025Set eng-2025-04264 marksMCQ
$\Omega$
›Reveal solutionSolution

Cutting the wire into $n$ equal pieces makes each $25/n\ \Omega$ ; connecting $n$ of them in parallel gives $25/n^2 = 1\ \Omega$ , so $n = 5$ .

Resistance of each piece:
$R_{piece} = \dfrac{25}{n}\ \Omega$
Parallel of $n$ equal resistors: …
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KEAM 2025Set eng-2025-04274 marksMCQ
$cm^2$
›Reveal solutionSolution

$R=\rho L/A$ with $L=2000$ m and $A=20\ \text{cm}^2=2\times10^{-3}\ \text{m}^2$ gives $R=10^{6}\rho$ .

Resistance of a uniform conductor is

$R=\rho\frac{L}{A}.$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$Q. If both the length and area of cross-section of a linear conductor are halved, its resistance would (A) be doubled (B) remain unchanged (C) be halved (D) be tripled (E) be quadrupled$
›Reveal solutionSolution

Resistance $R=\rho L/A$ . Halving the length and the area gives $R'=\rho(L/2)/(A/2)=\rho L/A=R$ — unchanged.

The resistance of a conductor is $R=\dfrac{\rho L}{A}$ . If both $L$ and $A$ are halved, $R'=\dfrac{\rho(L/2)}{(A/2)}=\dfrac{\rho L}{A}=R$ . The tw …
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KEAM 2024Set eng-2024-06064 marksMCQ
$^\circ$
›Reveal solutionSolution

Resistance rises linearly with temperature: $R=R_0(1+\alpha T)$ .

Using $R=R_0(1+\alpha T)$ with $R_0=4\ \Omega$ , $\alpha=5\times10^{-3}\,/^{\circ}$ C and $T=50\,^{\circ}$ C: …
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KEAM 2024Set eng-2024-06074 marksMCQ
$Q. Masses of three copper wires are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. Then the ratio of their electric resistances is (A) 125 : 15 : 1 (B) 5 : 3 : 1 (C) 1 : 25 : 125 (D) 1 : 3 : 5 (E) 5 : 21 : 25$
›Reveal solutionSolution

Resistance of a wire of fixed material is $R\propto L^2/m$ , giving $25:3:0.2 = 125:15:1$ .

Resistance $R=\dfrac{\rho L}{A}$ . The cross-sectional area is $A=\dfrac{\text{volume}}{L}=\dfrac{m/d}{L}$ , where $d$ is the (common) density. Hence $R=\dfrac{\rho L}{m/(dL)}=\dfrac{\rho d\,L^2}{m}$ , i.e. $R\propto \dfrac{L^2}{m}$ . …
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KEAM 2024Set pha-2024-06104 marksMCQ
$Q. Resistivity of a conductor increases with (A) increase in its length (B) decrease in its length (C) increase in its area of cross-section (D) decrease in its area of cross-section (E) increase in its temperature$
›Reveal solutionSolution

Resistivity is a material property, independent of geometry.

Resistivity $\rho$ depends on the material and temperature, not on length or cross-section (those affect resistance $R = \rho l/A$ , not $\rho$ ). For a metallic conductor, $\rho$ increases with temperature due to inc …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$Q. The INCORRECT statement is (A) Resistivity of copper increases with increase of temperature (B) Resistivity of germanium decreases with the increase of temperature (C) Resistivity of semiconductors is higher than that of the conductors (D) Resistivity of nichrome shows a weak dependence with temperature (E) Resistivity of insulators is independent of temperature$
›Reveal solutionSolution
The false statement is that insulators' resistivity is independent of temperature; it actually varies (decreasing) with temperature.

Concept and Intuition

Resistivity's temperature behaviour differs by material type: metals (copper, nichrome) increase with temperature; semiconductors (germanium) decrease with temperature; insulators are not temperature-independent — like semiconductors, their resistivity falls with rising temperature as more carriers become available.

Step-by-Step Solution

(A) Copper is a metal: resistivity increases with temperature — correct.

(B) Germanium is a semiconductor: resistivity decreases with temperature — correct.

(C) Semiconductors have higher resistivity than conductors — correct.

(D) Nichrome (alloy) has a weak temperature dependence — correct. …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$Q. Material that is widely used to make wire bound standard resistors is (A) manganin (B) iron (C) copper (D) tungsten (E) germanium$
›Reveal solutionSolution
Standard wire-wound resistors are made of manganin because of its very low temperature coefficient of resistance.

Concept and Intuition

A standard resistor must keep its resistance essentially constant with temperature and time. Manganin (Cu–Mn–Ni alloy) has an extremely small temperature coefficient of resistivity, making its resistance stable — ideal for precision standard resistors. Copper, iron and tungsten have significant temperature coefficients; germanium is a semiconductor.

Step-by-Step Solution

Requirement: near-zero temperature coefficient for a stable standard. …
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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

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

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