Two temperature scales A and B are related by a straight-line graph on which temperature on scale A (in °A) is plotted on the vertical axis and temperature on scale B (in °B) on the horizontal axis. Between the lower fixed point and the upper fixed point there are 150 equal divisions on scale A and 100 equal divisions on scale B. The straight line passes through the lower fixed point at $30\,^\circ$A (which reads $0\,^\circ$B) and the upper fixed point at $180\,^\circ$A (which reads $100\,^\circ$B). The relationship for conversion between the two scales is given by (a) $\dfrac{t_A - 180}{100} = \dfrac{t_B}{150}$ (b) $\dfrac{t_A - 30}{150} = \dfrac{t_B}{100}$ (c) $\dfrac{t_B - 180}{150} = \dfrac{t_A}{100}$ (d) $\dfrac{t_B - 40}{100} = \dfrac{t_A}{180}$

> [!TLDR] > A reading on one linear temperature scale converts to another by matching the *fraction of the way* each reading lies between the two scales' common fixed points. Here the lower fixed point is $30\,^\circ$A $= 0\,^\circ$B and the upper fixed point is $180\,^\circ$A $= 100\,^\circ$B, with 150 divisions on A and 100 on B between them. This gives $\dfrac{t_A-30}{150}=\dfrac{t_B}{100}$ — option (B). ### Concept Both scales are linear (equally spaced divisions), so plotting $t_A$ against $t_B$ is a straight line. Any two linear scales are related by matching the fraction of the interval between their common fixed points. ### Reading the graph - Lower fixed point (bottom of the line): $t_A = 30\,^\circ$A, $t_B = 0\,^\circ$B. - Upper fixed point (top of the line): $t_A = 180\,^\circ$A, $t_B = 100\,^\circ$B. - Interval on A: $180 - 30 = 150$ divisions; interval on B: $100 - 0 = 100$ divisions. ### Why this formula For a linear scale a reading $t$ is fixed by the fraction $f = \dfrac{t - (\text{lower fixed point})}{\text{number of divisions}}$. The *same* physical temperature must give the *same* fraction on both scales: $$\frac{t_A - 30}{150} = \frac{t_B - 0}{100}.$$ ### Steps 1. Scale A: fraction $= \dfrac{t_A - 30}{150}$. 2. Scale B: fraction $= \dfrac{t_B - 0}{100} = \dfrac{t_B}{100}$. 3. Equate the two fractions: $\dfrac{t_A - 30}{150} = \dfrac{t_B}{100}$. ### Why the distractors are wrong - **(A)** uses $t_A - 180$ (subtracts the *upper* fixed point) and swaps the divisions $100/150$ — both errors. - **(C)** subtracts 180 from $t_B$, but the fixed point $180$ belongs to scale A, not B; the axes/divisions are interchanged. - **(D)** contains $t_B - 40$ and a division of $180$, neither of which appears on either scale. > [!ANSWER] > Option **(B)**: $\dfrac{t_A - 30}{150} = \dfrac{t_B}{100}$.

Two temperature scales A and B are related by a straight-line graph on which temperature on scale A (in °A) is

NCERT Exemplar · Q3

30\,^\circ

(a)

\dfrac{t_A - 180}{100} = \dfrac{t_B}{150}

(b)

\dfrac{t_A - 30}{150} = \dfrac{t_B}{100}

(c)

\dfrac{t_B - 180}{150} = \dfrac{t_A}{100}

(d)

\dfrac{t_B - 40}{100} = \dfrac{t_A}{180}

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Concept understanding — Temperature Scale Conversion

Temperature Scale Conversion

You already know temperature as a measure of hotness or coldness. But here's the thing: different countries and different scientific fields measure that same hotness using different numbers. Water boils at 100 on the Celsius scale, but at 212 on the Fahrenheit scale. That's not because the water is different — it's because the rulers are different.

The core idea

A temperature scale is just a number line someone decided to draw on the same physical reality. Converting between scales means finding the number on one number line that corresponds to the same physical hotness as a given number on another number line.

Think of it like this: if you measure a table's length in feet and get 6, and I measure it in metres and get 1.83, we're both describing the same table. Temperature conversion works exactly the same way — same physical state, different numbers.

The three main scales you'll meet

Scale	Freezing point of water	Boiling point of water	Used by
Celsius (°C)	0	100	Most of the world, science
Fahrenheit (°F)	32	212	USA, a few other countries
Kelvin (K)	273.15	373.15	All scientific work

Notice something: Celsius and Kelvin have the same step size — a change of 1°C is exactly a change of 1 K. They just start at different places. Fahrenheit has a smaller step size (180 steps between freezing and boiling, instead of 100).

The conversion formulas

°F = \frac{9}{5}(°C) + 32

These aren't magic. Each one comes from the simple idea of matching two number lines.

Why the formulas look like that

Between freezing and boiling water:

Celsius has 100 steps (0 to 100)
Fahrenheit has 180 steps (32 to 212)

So one Celsius step is $\frac{180}{100} = \frac{9}{5}$ Fahrenheit steps. That's the $\frac{9}{5}$ factor. The $+32$ just shifts the starting point — because 0°C doesn't correspond to 0°F, it corresponds to 32°F.

For Kelvin: since the step size is identical to Celsius, you just add the offset. The 273.15 comes from the fact that absolute zero (the coldest possible temperature) is 0 K, which is $-273.15$ °C.

A worked example

Convert 25°C to Fahrenheit.

Step 1: Multiply by $\frac{9}{5}$ .

25 \times \frac{9}{5} = 25 \times 1.8 = 45

Step 2: Add 32.

45 + 32 = 77

So 25°C = 77°F. A warm spring day in Celsius terms is 77°F — same weather, different number.

Tip

For quick mental estimates: double the Celsius and add 30. It's not exact (you get 80 instead of 77 here), but it's close enough for everyday "should I wear a jacket?" decisions.

The one thing students mess up

Watch out

Never just add or subtract without the factor. "It's 30°C outside, so it must be 30 + 32 = 62°F" is wrong. You must multiply by $\frac{9}{5}$ first. The 32 is a shift after scaling, not before.

Why Kelvin matters

Kelvin is the scientist's scale because it starts at absolute zero — the point where particles have minimum possible thermal motion. This makes all gas laws and thermodynamic equations simple. You'll never see a negative Kelvin temperature in normal physics; 0 K is the floor.

When a problem gives you Celsius and the formula uses Kelvin (like the ideal gas law $PV = nRT$ ), you must convert: $K = °C + 273.15$ .

The big picture

Temperature scale conversion is just relabelling the same physical reality. The formulas are linear — multiply to adjust step size, add to adjust zero point. Once you see that, every conversion is just arithmetic.

For quick revision, remember that Temperature Scale Conversion is drawn directly from the Thermal Properties of Matter coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers, which is exactly why "Temperature Scale Conversion important questions" shows up so often in Physics question banks. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.

A reading on one linear temperature scale converts to another by matching the fraction of the way each reading lies between the two scales' common fixed points. Here the lower fixed point is $30\,^\circ$ A $= 0\,^\circ$ B and the upper fixed point is $180\,^\circ$ A $= 100\,^\circ$ B, with 150 divisions on A and 100 on B between them. This gives $\dfrac{t_A-30}{150}=\dfrac{t_B}{100}$ — option (B).

Concept

Both scales are linear (equally spaced divisions), so plotting $t_A$ against $t_B$ is a straight line. Any two linear scales are related by matching the fraction of the interval between their common fixed points.

Reading the graph

Lower fixed point (bottom of the line): $t_A = 30\,^\circ$ A, $t_B = 0\,^\circ$ B.
Upper fixed point (top of the line): $t_A = 180\,^\circ$ A, $t_B = 100\,^\circ$ B.
Interval on A: $180 - 30 = 150$ divisions; interval on B: $100 - 0 = 100$ divisions.

Why this formula

For a linear scale a reading $t$ is fixed by the fraction $f = \dfrac{t - (\text{lower fixed point})}{\text{number of divisions}}$ . The same physical temperature must give the same fraction on both scales:

\frac{t_A - 30}{150} = \frac{t_B - 0}{100}.

Steps

Scale A: fraction $= \dfrac{t_A - 30}{150}$ .
Scale B: fraction $= \dfrac{t_B - 0}{100} = \dfrac{t_B}{100}$ .
Equate the two fractions: $\dfrac{t_A - 30}{150} = \dfrac{t_B}{100}$ .

Why the distractors are wrong

(A) uses $t_A - 180$ (subtracts the upper fixed point) and swaps the divisions $100/150$ — both errors.
(C) subtracts 180 from $t_B$ , but the fixed point $180$ belongs to scale A, not B; the axes/divisions are interchanged.
(D) contains $t_B - 40$ and a division of $180$ , neither of which appears on either scale.

✓Final answer

\dfrac{t_A - 30}{150} = \dfrac{t_B}{100}

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

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$-20^0C$
›Reveal solutionSolution
Setting the Celsius and Fahrenheit readings equal in the conversion formula gives a unique coincidence point at $-40^\circ$ (on both scales).

Concept and Intuition

The Celsius and Fahrenheit scales are two different linear scales for the same physical quantity (temperature), related by an affine (not just proportional) transformation. Because the transformation is affine ( $F = \frac{9}{5}C+32$ , with a nonzero slope different from 1), the two scales intersect (read the same number) at exactly one point.

Step-by-Step Solution

Conversion formula: $F = \dfrac{9}{5}C + 32$ .

We want the temperature $x$ at which the numerical reading is the same on both scales: set $C = F = x$ .

$x = \dfrac{9}{5}x + 32$ .

$x - \dfrac{9}{5}x = 32 \Rightarrow -\dfrac{4}{5}x = 32$ .

$x = 32 \times \left(-\dfrac{5}{4}\right) = -40$ .

So at $-40^\circ$ C the Fahrenheit reading is also $-40^\circ$ F — this is the well-known coincidence point.

Common Mistakes

Sign errors when moving terms across the equation.

Confusing this coincidence point with the point where Kelvin and Celsius numerically relate (that's a different, additive-only, relationship — $K = C + 273$ , which never coincides for any finite value).

✓Final answer
$-40^\circ\text{C}$

ANSWER: B
AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$50^{\circ}\text{C}$
›Reveal solutionSolution
Tests the difference between converting a temperature value and converting a temperature interval/change between Celsius and Fahrenheit.

Concept and Intuition

The formula $F = \tfrac95 C + 32$ converts an absolute reading. But when converting a change in temperature ( $\Delta T$ ), the $+32$ offset cancels out (it appears on both the initial and final readings and subtracts away), leaving only the scale-factor $\tfrac95$ (or $\tfrac59$ the other way). This is a very common trap: treating an "increase of 54°F" as if it means "the new reading is 54°F."

Step-by-Step Solution

The body starts at $50^{\circ}\text{C}$ and its temperature is raised by $54^{\circ}\text{F}$ — this 54°F is an interval, not a final reading.

Convert the interval to Celsius: since $1^{\circ}\text{C}$ change $=\tfrac95{}^{\circ}\text{F}$ change, a $54^{\circ}\text{F}$ change $=54\times\tfrac{5}{9}=30^{\circ}\text{C}$ change.

New Celsius temperature: $50+30 = 80^{\circ}\text{C}$ .

Convert this final reading to Fahrenheit (now the $+32$ offset does apply, since this is an absolute value): $F=\tfrac95(80)+32=144+32=176^{\circ}\text{F}$ .

Matching against the options, $176^{\circ}\text{F}$ is listed — option (D).

Common Mistakes

Adding $54^{\circ}\text{F}$ directly onto $50^{\circ}\text{C}$ as if they were the same unit.

Applying the $+32$ offset while converting the interval itself (it should only be used when converting an absolute reading).

✓Final answer
$176^{\circ}\text{F}$

ANSWER: D
AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Q. A Celsius and a Fahrenheit thermometers are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers 131°. The falls in temperature as registered by the Celsius thermometer is (A) 45° (B) 40° (C) 60° (D) 75°$
›Reveal solutionSolution
Converting the final Fahrenheit reading (131°F) to Celsius (55°C) and subtracting from the initial 100°C boiling point gives a fall of 45°C. Answer: (A).

Concept and Intuition

Both thermometers read the same physical temperature at every instant — they simply use different scales. So to compare a fall as measured on the Celsius scale, first convert the final Fahrenheit reading into its equivalent Celsius value using the standard linear conversion, then simply subtract from the known starting Celsius temperature (100°C, boiling water at standard pressure).

Step-by-Step Solution

Initial temperature (boiling water): $100^\circ C$ (which is $212^\circ F$ ).

Final Fahrenheit reading: $131^\circ F$ . Convert: $C = \dfrac{5}{9}(F-32) = \dfrac{5}{9}(131-32) = \dfrac{5}{9}\times99 = 55^\circ C$ .

Fall in Celsius reading $= 100^\circ C - 55^\circ C = 45^\circ C$ .

Common Mistakes

Using the conversion formula backwards (converting Celsius to Fahrenheit) instead of Fahrenheit to Celsius.

Forgetting to subtract 32 before scaling by $5/9$ .

✓Final answer
$The correct option is (A) — 45°.$

ANSWER: A
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$13:5$
›Reveal solutionSolution
Solving $F/C = 13/5$ with the Celsius-Fahrenheit relation gives $C=40\,^0\mathrm C$ (equivalently $F=104\,^0\mathrm F$ ); the option correctly labelled is $40\,^0\mathrm C$ .

Concept and Intuition

The Celsius and Fahrenheit scales are linearly related by $F=\tfrac95C+32$ . Given a ratio between the two numeric readings (not a difference), substitute the relation and solve the resulting linear equation in one unknown.

Step-by-Step Solution

Let the Celsius reading be $C$ and Fahrenheit reading be $F=\tfrac95C+32$ .

Given $\dfrac{F}{C}=\dfrac{13}{5}\Rightarrow 5F=13C$ .

Substitute: $5\left(\dfrac95C+32\right)=13C \Rightarrow 9C+160=13C$ .

$160=4C \Rightarrow C=40$ , and $F=\tfrac95(40)+32=72+32=104$ .

Check ratio: $104:40=13:5$ ✓. So the temperature is $40\,^0\mathrm C$ (equal to $104\,^0\mathrm F$ ); among the options only " $40\,^0\mathrm C$ " is correctly labelled.

Common Mistakes

Mislabeling the Fahrenheit value (104) as Celsius, which is a trap option.

Using the ratio as a difference ( $F-C=13-5$ ) instead of $F/C=13/5$ .

✓Final answer
$40\ {}^0\mathrm{C}$

ANSWER: C
AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$Q. If two temperatures on Celsius scale differ by 25°, then the difference of those two temperatures on Fahrenheit scale is (A) 45° (B) 25° (C) 32° (D) 59°$
›Reveal solutionSolution
Converting a difference of temperature from Celsius to Fahrenheit uses only the scale-factor $9/5$ — the $32°$ offset cancels out — giving $45°$ .

Concept and Intuition

The Celsius-to-Fahrenheit conversion is $F = \dfrac95C+32$ . When we look at a difference between two readings, $\Delta F = F_2-F_1 = \dfrac95(C_2-C_1) = \dfrac95\Delta C$ — the additive constant $32$ drops out because it's the same in both readings.

Step-by-Step Solution

Given $\Delta C = 25°$ .

$\Delta F = \dfrac95\times\Delta C = \dfrac95\times25 = 45°$ .

Common Mistakes

Adding 32° to the converted difference (that's only valid for converting an absolute reading, not a difference).

Using the reciprocal factor $5/9$ instead of $9/5$ .

✓Final answer
$The correct option is (A) — 45°.$

ANSWER: A
AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$Q. The nature of the graph between the temperature in Fahrenheit and Celsius values is (A) a parabola (B) an exponential curve (C) a sine wave (D) a straight line$
›Reveal solutionSolution
The Fahrenheit–Celsius conversion formula is a first-degree (linear) equation, so its graph is necessarily a straight line.

Concept and Intuition

Temperature scales like Celsius and Fahrenheit are both linear scales built on fixed reference points (freezing/boiling points of water), related by an affine (linear) transformation: $F = \dfrac{9}{5}C+32$ . Any equation of the form $y = mx+c$ (constant slope $m$ , constant intercept $c$ ) graphs as a straight line, since equal increments in $x$ always produce equal increments in $y$ .

Step-by-Step Solution

Recall the conversion relation: $F = \dfrac{9}{5}C + 32$ .

This is of the form $y = mx + c$ with $m = 9/5$ (a constant slope) and $c = 32$ (a constant intercept) — the defining form of a straight line.

Since neither $m$ nor $c$ depends on $C$ or $F$ , and there are no higher powers, exponentials, or periodic terms, the C–F graph must be linear, i.e., a straight line.

Common Mistakes

Overthinking the question by looking for exotic behaviour; the relation is a simple linear equation, so the graph is a straight line, not a curve.

✓Final answer
$The correct option is (D) — a straight line.$

ANSWER: D
AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$-183\ ^\circ C$
›Reveal solutionSolution
A straightforward Celsius-to-Fahrenheit conversion: $-183^\circ C$ converts to approximately $-297^\circ F$ .

Concept and Intuition

The Celsius and Fahrenheit scales are linearly related, with $0^\circ C = 32^\circ F$ and a $100$ -degree Celsius span mapping to a $180$ -degree Fahrenheit span (ratio $9/5$ ). Converting simply means applying this linear relation.

Step-by-Step Solution

Conversion formula: $F = \dfrac{9}{5}C + 32$ .

Substitute $C = -183$ :

$F = \frac{9}{5}\times(-183) + 32 = -329.4 + 32$

Compute:

$F = -297.4^\circ F \approx -297^\circ F$
Common Mistakes

Using $F = \dfrac{5}{9}C+32$ (inverting the conversion ratio) — that formula is for converting Fahrenheit to Celsius, not the reverse.

Forgetting to add the $32$ offset after scaling.

✓Final answer
$-297\ ^\circ F$

ANSWER: B

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Concept

Reading the graph

Why this formula

Steps

Why the distractors are wrong

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