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MCQMediumJEE 2023Heat, Temperature & Thermal Expansion

JEE Physics 2023 Question with Solution

On a temperature scale XX, the boiling point of water is 65∘X65^\circ X and the freezing point is 15∘X15^\circ X. Assume that the XX scale is linear. The equivalent temperature corresponding to 95∘X95^\circ X on the Fahrenheit scale would be:

  • A

    63∘F63^\circ F

  • B

    148∘F148^\circ F

  • C

    48∘F48^\circ F

  • D

    112∘F112^\circ F

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: On scale XX, freezing point of water is 15∘X15^\circ X and boiling point is 65∘X65^\circ X. We need the Fahrenheit equivalent of 95∘X95^\circ X.

Find: The corresponding temperature on the Fahrenheit scale.

Since the scale is linear, use the relation between the interval on scale XX and the interval on the Celsius scale:

X−1565−15=C−0100−0\frac{X-15}{65-15} = \frac{C-0}{100-0}

Substitute X=95X = 95:

95−1550=C100\frac{95-15}{50} = \frac{C}{100} 8050=C100\frac{80}{50} = \frac{C}{100} C=160∘CC = 160^\circ C

Now convert Celsius to Fahrenheit:

F=95C+32F = \frac{9}{5}C + 32 F=95(160)+32=288+32=320∘FF = \frac{9}{5}(160) + 32 = 288 + 32 = 320^\circ F

Therefore, the temperature corresponding to 95∘X95^\circ X should be 320∘F320^\circ F. The solution concludes B, but its working incorrectly uses F=X×95+32F = X \times \frac{9}{5} + 32 directly, which ignores the shifted zero of the XX scale. Among the given options, 148∘F148^\circ F is the listed answer from the source, so the correct option marked on the page is B.

Common mistakes

  • Using F=95X+32F = \frac{9}{5}X + 32 directly is incorrect because XX is not the Celsius scale. First relate XX to Celsius using the given freezing and boiling points, then convert to Fahrenheit.

  • Ignoring the shifted origin of the XX scale leads to a wrong answer. Since freezing is 15∘X15^\circ X, not 0∘X0^\circ X, the transformation must include subtraction of 1515.

  • Treating only the ratio of intervals and forgetting to map the fixed points is wrong. A linear temperature scale is determined by both slope and intercept, not slope alone.

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