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⚛️ NEET Physics❓ Question

Faulty thermometer reads 5° at ice point and 99° at steam point, how to find the real temperature?

Rohan Sharma11th Sample

24 Sept 2026 · 1 min read

ok so thermal properties of matter and the first page is "temperature scales" and I'm already lost lol

Q: a faulty thermometer reads 5° at the ice point and 99° at the steam point. When it reads 52°, what's the actual temperature? I did 52−599×100=47.5\dfrac{52-5}{99}\times100 = 47.5 and the options are 45, 50, 52 and 99. 47.5 is not there 💀

Also I forget the conversion between Celsius and Fahrenheit: is it F=95C+32F = \frac95 C + 32 or the other way?

And one question said: "the temperature of a body rises by 20°C. What is the rise on the Kelvin and Fahrenheit scales?" I wrote 293 K and 68°F. Bro what.

I know heat is energy and temperature is "how hot". That's about it. How should I think about all these scales so I don't have to memorise 10 formulas?

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  • [+]Ishita GoyalDropperSample· 25d ago

    Your mistake: you divided by 99, but the full range of the faulty thermometer is 99−5=9499 - 5 = 94 divisions, not 99. Use the ratio form:

    reading−LFPUFP−LFP=C−0100−0\frac{\text{reading} - \text{LFP}}{\text{UFP} - \text{LFP}} = \frac{C - 0}{100 - 0}

    52−599−5=4794=12=C100⇒C=50°C\frac{52-5}{99-5} = \frac{47}{94} = \frac12 = \frac{C}{100} \Rightarrow C = 50°\text{C}

    Same trick for any two scales. Think of the scales as two rulers with different zeros and different unit lengths: the fraction of the way from freezing to boiling is the same on both.

    • [+]Rohan Sharma11thSample· 25d agoAuthor

      the 94 not 99 thing, I'm so done with myself 😭 thanks. will do the ratio method always

  • [+]Yash Tiwari11thSample· 25d ago

    Hinglish memory hack: 0°C = 32°F and 100°C = 212°F, 180 divisions on the F scale for 100 on the C scale, so each °C is 1.8 °F. Body temperature 37°C: 95×37+32=98.6\tfrac95\times37 + 32 = 98.6°F, which is the number on every thermometer. Isi se formula yaad rakh lo, ratio waali.

  • [+]Farhan ShaikhDropperSample· 25d ago

    On your third one (the 20°C rise) that's the common trap. The "+273" and "+32" are only for converting an actual temperature reading. A change in temperature has no offset. So

    • ΔK = ΔC = 20 K
    • ΔF = (9/5) × 20 = 36°F

    If a body is at 20°C, then it's 293 K and 68°F. But if it rises by 20°C, it rises by 20 K and 36°F. 293 K and 68°F is exactly the mix-up.

    • [+]Rohan Sharma11thSample· 25d agoAuthor

      ohh offset only for readings, not for changes. yep that's the trap I fell in. thanks Farhan

  • [+]Lakshmi TripathiDropperSample· 25d ago

    Conversion in the ratio form, which is easier to hold on to:

    C5=F−329=K−2735\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273}{5}

    So F=95C+32F = \frac95C + 32 (what you first wrote is right). At −40-40 the Celsius and Fahrenheit scales have the same reading. Bonus: it's one of those one-mark MCQs.

  • [+]Siddharth RaoDropperSample· 25d ago

    On "what even is temperature". A thermometer works because some property changes with temperature: length of mercury, resistance of a wire, pressure of a gas at constant volume. The constant-volume gas thermometer is the standard reference, because for low-density gases P∝TP \propto T and the readings agree whatever gas you use. The Kelvin scale is fixed using the triple point of water, 273.16 K. Celsius is just Kelvin shifted by 273.15.

    Heat is energy flowing because of a temperature difference. Temperature isn't "amount of heat". Two things at the same temperature don't exchange heat. That's the idea behind the zeroth law, you'll meet it soon.

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