The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: $$R = R_0\left[1 + \alpha (T - T_0)\right]$$ The resistance is $101.6\ \Omega$ at the triple-point of water $273.16\ \text{K}$, and $165.5\ \Omega$ at the normal melting point of lead ($600.5\ \text{K}$). What is the temperature when the resistance is $123.4\ \Omega$?

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximat

Exercises · 10.3

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

Karnataka PUCTextbookSubjective· 3mImportance★★★★★est

20% · 11/55 Questions

✓ Free question

Concept understanding — Linear Temperature Dependence

Linear Temperature Dependence – First Encounter

You've probably noticed that many things change when you heat them up. A metal rail expands on a hot day. The resistance of a wire increases when current makes it hot. The pressure in a sealed tyre rises after a long drive.

The simplest way this happens is linear temperature dependence — the property changes in direct proportion to the change in temperature. Double the temperature rise, double the change in the property. No surprises, no sudden jumps.

The Intuition

Imagine a rubber band. If you pull it gently, it stretches a little. Pull twice as hard, it stretches twice as much — that's a linear relationship between force and stretch.

Now replace "force" with "temperature change" and "stretch" with "some physical quantity" (length, resistance, pressure, volume). That's linear temperature dependence: equal increments of temperature produce equal increments of the quantity.

Note

This is the first approximation for most materials over a limited temperature range. It's never perfectly true forever, but it's remarkably accurate for small temperature changes.

The Precise Statement

If a physical quantity $Q$ depends linearly on temperature $T$ , then:

Q(T) = Q_0 + \alpha \, (T - T_0)

Where:

$Q_0$ is the value at some reference temperature $T_0$ (often $0^\circ\text{C}$ or $25^\circ\text{C}$ )
$\alpha$ is the temperature coefficient — the rate of change per degree
$T$ is the current temperature

The change $\Delta Q = Q - Q_0$ is directly proportional to the change $\Delta T = T - T_0$ :

\Delta Q = \alpha \, \Delta T

Q(T) = Q_0 \, [1 + \beta (T - T_0)]

Real Examples You'll Meet in Exams

Quantity	Symbol	Typical behaviour	Common $\beta$ value
Length of a metal rod	$L$	Expands on heating	$\approx 1.2 \times 10^{-5} \,^\circ\text{C}^{-1}$ (steel)
Resistance of a copper wire	$R$	Increases with temperature	$\approx 3.9 \times 10^{-3} \,^\circ\text{C}^{-1}$
Volume of an ideal gas (constant pressure)	$V$	Increases linearly with $T$ (in Kelvin)	$1/273.15 \,^\circ\text{C}^{-1}$
Pressure of an ideal gas (constant volume)	$P$	Increases linearly with $T$ (in Kelvin)	$1/273.15 \,^\circ\text{C}^{-1}$

Watch out

For gases, the linear law works only when temperature is measured in Kelvin, not Celsius. The formula becomes $V = V_0 (1 + \frac{T}{273.15})$ where $T$ is in °C — but this is just a disguised version of $V \propto T$ (Kelvin).

Why This Matters

Linear temperature dependence is the foundation of:

Thermometers (mercury in glass, resistance thermometers, thermocouples)
Thermal expansion calculations in bridges and railway tracks
Temperature compensation in electronic circuits
Charles's Law and Gay-Lussac's Law for ideal gases

The key insight: when you see a straight-line graph of a physical quantity against temperature, you're looking at linear temperature dependence. The slope of that line is $\alpha$ , and the intercept at $T=0$ (or $T=T_0$ ) gives you $Q_0$ .

Important

Linear temperature dependence is not a law of nature — it's an approximation that works well for small temperature ranges. For large temperature changes, higher-order terms ( $T^2$ , $T^3$ , ...) become important, and the relationship becomes non-linear.

Students preparing for boards often pair a search for "Linear Temperature Dependence class 11 physics" with "NCERT Physics syllabus" — Linear Temperature Dependence is a syllabus-aligned topic under Thermal Properties of Matter in NCERT Class 11 Physics, making it a natural fit for both board exams and JEE/NEET practice sets. Working through the worked examples above alongside the official NCERT Physics textbook is the most reliable way to turn this understanding into exam-ready recall.

Treating the resistance-temperature relation as linear between the two known calibration points, interpolation gives $T\approx384.9$ K (about $385$ K, or $111.7^\circ$ C) for a resistance of $123.4\ \Omega$ .

The relation $R=R_0[1+\alpha(T-T_0)]$ says resistance changes in direct proportion to temperature above the reference point $T_0=273.16$ K (the triple point of water), where $R_0=101.6\ \Omega$ . Since it's linear, the ratio of resistance changes equals the ratio of temperature changes between any two points on the line.

Setting up the ratio

We know two points on the line: $(273.16\ \text{K}, 101.6\ \Omega)$ and $(600.5\ \text{K}, 165.5\ \Omega)$ . For the unknown point $(T, 123.4\ \Omega)$ :

\frac{R-R_0}{R_1-R_0} = \frac{T-T_0}{T_1-T_0}.

Substituting values

\frac{123.4-101.6}{165.5-101.6} = \frac{T-273.16}{600.5-273.16}

\frac{21.8}{63.9} = \frac{T-273.16}{327.34}

Solving for $T$

T - 273.16 = \frac{21.8}{63.9}\times327.34 \approx111.7\ \text{K}

T \approx273.16 + 111.7 \approx384.9\ \text{K}.

Rounding to match the precision of the data, $T\approx385$ K, or equivalently about $111.7^\circ$ C.

Tip

This linear-interpolation method sidesteps solving for $\alpha$ explicitly - dividing the two known ratios directly is faster and less error-prone than computing $\alpha$ first and then re-solving.

✓Final answer

384.9

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

KCET 2025Set D-41 markMCQ
$5200\Omega$
›Reveal solutionSolution

Apply the standard half-deflection result $G = \dfrac{RS}{R - S}$ , with $R = 5200\ \Omega$ (the series resistance) and $S = 90\ \Omega$ (the shunt that halves the deflection).

Step 1 — What happens with $K_2$ open.

Only the series resistance $R$ and the galvanometer $G$ carry the current. If the cell emf is $E$ (internal resistance negligible), the current through the galvanometer is
$I_g = \frac{E}{R + G}$
This produces the full deflection, $\theta = 26$ divisions. Since deflection $\propto$ current through the coil,
$26 \propto \frac{E}{R+G} \qquad (1)$
Step 2 — What happens when $K_2$ is closed.

The shunt $S = 90\ \Omega$ is now in parallel with the galvanometer. Because $R \gg$ the parallel combination ( $5200\ \Omega$ versus at most $90\ \Omega$ ), the total current drawn from the cell is essentially unchanged:
$I \approx \frac{E}{R}$
That current now splits between $G$ and $S$ . By the current-divider rule the fraction through the galvanometer is
$I_g' = I \cdot \frac{S}{G + S}$
The observed deflection is halved, to $13$ divisions.

Step 3 — Impose the half-deflection condition.

Deflection halves $\Rightarrow$ the coil current halves:
$\frac{I_g'}{I_g} = \frac{13}{26} = \frac{1}{2}$ $\frac{S}{G+S} = \frac{1}{2}\ \text{(of the near-unchanged total)} \quad\Longrightarrow\quad \frac{E}{R}\cdot\frac{S}{G+S} = \frac{1}{2}\cdot\frac{E}{R+G}$
Working this through (the standard derivation) yields the well-known result
$\boxed{G = \frac{RS}{R - S}}$
Why this form? In the ideal limit the shunt must carry exactly as much current as the coil for the coil current to halve — which requires $S = G$ if the total were fixed. The finite $R$ correction produces the $R/(R-S)$ factor, and because $R \gg S$ that factor is only slightly above 1, so $G$ comes out slightly larger than $S$ . We should therefore expect an answer a little above $90\ \Omega$ — a useful check before computing.

Step 4 — Substitute the numbers.
$G = \frac{5200 \times 90}{5200 - 90} = \frac{468000}{5110}$ $G = 91.585\ldots \approx 91.6\ \Omega$
This sits just above $S = 90\ \Omega$ , exactly as anticipated.

Rejecting the distractors: (A) $45.0\ \Omega$ is $S/2$ (halving the wrong quantity); (B) $103\ \Omega$ and (D) $116\ \Omega$ do not follow from $RS/(R-S)$ with these values.

✓Final answer
$91.6\ \Omega$

ANSWER: C
KCET 2019Set A-11 markMCQ
$Q. Masses of three wires of copper are in the ratio 1:3:5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are (A) 1:3:5 (B) 5:3:1 (C) 1:15:125 (D) 125:15:1$
›Reveal solutionSolution
Resistance depends on length and cross-sectional area; using mass and density to relate area gives the ratio of resistances as $125:15:1$ , which corresponds to option (D).

The key idea here is that resistance $R$ of a wire is given by $R = \rho \frac{L}{A}$ , where $\rho$ is resistivity (same for all wires since they are all copper), $L$ is length, and $A$ is cross-sectional area. We are not given areas directly, but we are given masses and lengths. Since all wires are made of the same material (copper), their densities are equal. Mass $m = \text{density} \times \text{volume} = d \times (A \times L)$ . So $A = \frac{m}{d L}$ . Substituting this into the resistance formula lets us express $R$ in terms of $m$ and $L$ alone.

Let’s work through it step by step.

Write the resistance formula in terms of mass and length. For a wire of resistivity $\rho$ , length $L$ , and cross-sectional area $A$ :

$R = \rho \frac{L}{A}$
Since mass $m = d \cdot A \cdot L$ (where $d$ is density), we have $A = \frac{m}{d L}$ . Substitute:
$R = \rho \frac{L}{\frac{m}{d L}} = \rho \frac{L \cdot d L}{m} = \frac{\rho d \, L^2}{m}$
Because $\rho$ and $d$ are constants for all wires, $R \propto \frac{L^2}{m}$ .

Assign the given ratios as actual numbers.

The masses are in the ratio $1:3:5$ . Let the masses be $m_1 = 1k$ , $m_2 = 3k$ , $m_3 = 5k$ (where $k$ is some constant).

The lengths are in the ratio $5:3:1$ . Let the lengths be $L_1 = 5\ell$ , $L_2 = 3\ell$ , $L_3 = 1\ell$ .

Compute the resistance ratio.

Since $R \propto \frac{L^2}{m}$ , we can write:

$R_1 : R_2 : R_3 = \frac{L_1^2}{m_1} : \frac{L_2^2}{m_2} : \frac{L_3^2}{m_3}$
Substitute the values:
$= \frac{(5\ell)^2}{1k} : \frac{(3\ell)^2}{3k} : \frac{(1\ell)^2}{5k}$ $= \frac{25\ell^2}{k} : \frac{9\ell^2}{3k} : \frac{1\ell^2}{5k}$
Simplify each term:
$= 25 : 3 : \frac{1}{5}$

Eliminate fractions to get a clean integer ratio. Multiply each term by 5:

$= 125 : 15 : 1$
Watch out

A common mistake is to directly use $R \propto L/A$ and forget that area itself depends on mass and length. Another pitfall is inverting the mass ratio — since $R \propto 1/m$ , larger mass means smaller resistance, so the smallest mass gives the largest resistance.

Tip

Notice that the ratio $L^2/m$ quickly gives the answer without needing to compute actual areas. This shortcut works whenever the material (and hence density and resistivity) is the same for all wires.

✓Final answer
$125:15:1$
KCET 2018Set A-11 markMCQ
$r_1$
›Reveal solutionSolution

Write the series current from the loop equation, then impose the condition 'terminal PD of cell 1 $= 0$ ', i.e. $V_1 = E - Ir_1 = 0$ .

Step 1 — Current in the series loop.

Two cells of the same emf $E$ in series drive a total emf $2E$ against the total resistance $R + r_1 + r_2$ :
$I = \frac{2E}{R + r_1 + r_2}$
Step 2 — Terminal potential difference of a cell.

A cell delivering current $I$ has terminal PD
$V = E - I r$
because the current drops $Ir$ volts inside the cell across its own internal resistance. For the first cell:
$V_1 = E - I r_1$
Step 3 — Apply the given condition $V_1 = 0$ .
$E - I r_1 = 0 \;\Rightarrow\; E = I r_1$
Physically: the whole emf of cell 1 is consumed internally — it is being 'short-circuited' by the rest of the circuit, so it delivers no useful voltage.

Step 4 — Substitute the current.
$E = \frac{2E}{R + r_1 + r_2}\; r_1$
Cancel $E$ (non-zero):
$R + r_1 + r_2 = 2 r_1$ $\boxed{R = r_1 - r_2}$
Sanity check: this requires $r_1 > r_2$ — reasonable, since only the cell with the larger internal resistance can have its entire emf eaten up internally.

✓Final answer
$R = r_1 - r_2$

ANSWER: C

←Previous question10.2 — Two absolute scales A and B have triple points of water defined to be …→Next question10.4 — Answer the following: (a) The triple-point of water is a standard fixe…

Setting up the ratio

Substituting values

Solving for TTT

🎓Unlock everything free for 14 days

Solving for $T$

Unlock everything free for 14 days