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.

Showing the 12 most recent of 23 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The boiling point of water is ______ Celsius at 1.013 bar pressure.$
›Reveal solutionSolution

Water's normal boiling point, at standard atmospheric pressure of 1.013 bar, is 100°C (373.15 K).

The boiling point of a liquid is the temperature at which its vapour pressure equals the external (atmospheric) pressure. This value depends on the surrounding pressure — at lower pressures (e.g., high altitude) water boils at a lower temperature, and at higher pressures it boils at a higher temperature. The standard atmospheric pressure is defined as 1.013 bar (1 atm, or 101.325 kPa), and at this pressure the boiling point of pure water is exactly 100°C.

✓Final answer
$The boiling point of water at 1.013 bar pressure is 100°C.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The temperature at which Celsius temperature and Fahrenheit temperature are equal, is$
(a) 0°
(b) 40°
(c) − 40°
(d) − 32°
›Reveal solutionSolution

Celsius and Fahrenheit are equal at -40 degrees. Answer (C).

The conversion is F = (9/5)C + 32. Set F = C = x:

x = (9/5)x + 32

x - (9/5)x = 32

-(4/5)x = 32

x = -40.

So at -40, the Celsius and Fahrenheit readings are the same (-40 C = -40 F).

✓Final answer
$(C) -40 degrees.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The temperature which have same numerical value in Celsius and Fahrenheit scales is (A) 40° (B) -40°C (C) 100° (D) 273°$
›Reveal solutionSolution

The Celsius and Fahrenheit scales read the same numerical value at -40°.

The relation between Celsius (C) and Fahrenheit (F) scales is:

$\dfrac{C}{5} = \dfrac{F-32}{9}$

Setting $C = F$ (equal numerical value) and solving:

$\dfrac{C}{5} = \dfrac{C-32}{9} \Rightarrow 9C = 5C - 160 \Rightarrow 4C = -160 \Rightarrow C = -40$

So at $-40°$ , both scales show the same reading: $-40°C = -40°F$ .

✓Final answer
$(B) -40°C.$
CBSE 2025Set hz1 markMCQ
$Q. On the basis of absolute zero as the scale of temperature, the ice melts at:$
(a) 100 degrees
(b) 273.15 K
(c) 373.15 K
(d) None of them
›Reveal solutionSolution

The Kelvin scale is defined so that its zero coincides with absolute zero, and the ice point (0 degree C, where ice melts under standard pressure) is 273.15 K on this scale.

The absolute (Kelvin) temperature scale is related to the Celsius scale by:

T(K) = T(degree C) + 273.15

The melting point of ice (the ice point) under standard atmospheric pressure is 0 degree C by definition of the Celsius scale. Converting this to the Kelvin scale:

T(K) = 0 + 273.15 = 273.15 K

By contrast, 373.15 K corresponds to 100 degree C, the boiling point of water at standard pressure -- the value in option (c) is the steam point, not the ice point.

✓Final answer
$The correct option is (b) 273.15 K.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The temperature of a body is found in Kelvin scale as x K and in Fahrenheit scale x°F. The value of x will be (A) 574.25 (B) 301.25 (C) 335 (D) 365$
›Reveal solutionSolution

The temperature at which the Kelvin and Fahrenheit numerical values coincide is $x=574.25$ .

Conversion: $°F=\dfrac95(K-273)+32$ . Setting both readings equal to $x$ : $x=\dfrac95(x-273)+32$ .

Multiply by 5: $5x=9(x-273)+160=9x-2457+160=9x-2297$ .

$5x-9x=-2297 \Rightarrow -4x=-2297 \Rightarrow x=574.25$ .

✓Final answer
$(A) 574.25.$
CBSE 2024Set SET-AP55001 markMCQ
$Q. The relation between the Centigrade (Celsius) temperature scale and the Kelvin temperature scale is:$
(a) c = k - 273
(b) c = k + 273
(c) c = k + 100
(d) c = k - 100
›Reveal solutionSolution

The Celsius and Kelvin scales have the same size of degree but different zero points: 0°C = 273 K (more precisely 273.15 K), so C = K − 273.

The Kelvin scale starts at absolute zero (0 K), while the Celsius scale sets 0°C at the freezing point of water. The freezing point of water is 273.15 K, so to convert from Kelvin to Celsius you subtract 273 (approximately):

C = K − 273

Check: at K = 373 (boiling point of water), C = 373 − 273 = 100°C — correct. At K = 273, C = 0°C — correct.

✓Final answer
$The correct option is (a) c = k - 273.$
CBSE 2024Set SET-AP55001 mark
$Q. At what temperature will the readings of the Fahrenheit and Celsius scales be equal?$
›Reveal solutionSolution

The Fahrenheit and Celsius scales give the same numerical reading at exactly −40 degrees.

The conversion between the two scales is:

F = (9/5)C + 32

Setting F = C (both readings equal, call the common value x):

x = (9/5)x + 32

x − (9/5)x = 32

−(4/5)x = 32

x = 32 × (−5/4) = −40

So at −40°, the Celsius and Fahrenheit thermometers show the identical numerical value: −40°C = −40°F.

✓Final answer
$-40° (-40°C = -40°F).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The SI unit of temperature is$
(a) A) °C
(b) B) °F
(c) C) K
(d) D) Kg
›Reveal solutionSolution
[!TLDR]

C) K

Why

Kelvin is the SI base unit of thermodynamic temperature.

[!ANSWER]

C) K
CBSE 2023Set ANNUAL1 markMCQ
$Q. SI unit of temperature is$
(a) Kelvin
(b) Celsius
(c) Fahrenheit
(d) Centigrade
›Reveal solutionSolution

Kelvin is the SI base unit for temperature; Celsius, Fahrenheit and Centigrade are common but non-SI temperature scales.

The International System of Units (SI) defines seven base units, one for each fundamental physical quantity. For temperature, the SI base unit is the kelvin (K), defined via the thermodynamic (absolute) temperature scale, with 0 K = absolute zero. Celsius (°C) is an SI-accepted derived scale (T(K) = T(°C) + 273.15) but is not itself the SI base unit; Fahrenheit is not part of the SI system at all. "Centigrade" is simply an older name for the Celsius scale.

✓Final answer
$(a) Kelvin.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Temperature difference of 20°C is equivalent to:$
(a) 20 K
(b) 293 K
(c) 253 K
(d) 0 K
›Reveal solutionSolution

Celsius and Kelvin scales have identically sized degrees, so a temperature DIFFERENCE of 20 degC is exactly 20 K -- only absolute temperature readings differ by the 273 offset, not temperature intervals.

The relation between an absolute Celsius reading and Kelvin reading is T(K) = T(degC) + 273. For two temperatures T1 and T2 in Celsius, converting both to Kelvin:

T1(K) = T1(degC) + 273

T2(K) = T2(degC) + 273

Subtracting: T2(K) - T1(K) = T2(degC) - T1(degC)

So the DIFFERENCE is unchanged by the conversion -- the +273 offset cancels out. A temperature CHANGE of 20 degC is therefore also a change of 20 K. (293 K and 253 K would be relevant if this were an absolute temperature reading of 20 degC or -20 degC, not a difference.)

✓Final answer
$The correct option is (a) 20 K.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Absolute zero of temperature is:$
(a) 100°C
(b) 273°C
(c) −273°C
(d) −273.15°C
›Reveal solutionSolution

Absolute zero = −273.15 °C = 0 K.

The Kelvin (absolute) scale is related to Celsius by T(K) = t(°C) + 273.15. At absolute zero T = 0 K, so t = −273.15 °C. This is the lowest temperature theoretically attainable, at which molecular motion is minimum.

✓Final answer
$(D) -273.15°C.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Temprature which has equal reading on Celcius and Faherenhit scale:$
(a) 0°
(b) 30°
(c) −40°
(d) 40°
›Reveal solutionSolution

The two scales agree at −40°.

The Celsius–Fahrenheit relation is F = (9/5)C + 32. For equal readings put F = C = x:

x = (9/5)x + 32 ⇒ x − (9/5)x = 32 ⇒ −(4/5)x = 32 ⇒ x = −40.

So −40 °C = −40 °F.

✓Final answer
$(C) -40°.$

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Concept

Reading the graph

Why this formula

Steps

Why the distractors are wrong

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