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Question: Figure shows three temperature scales with the freezing and boiling point of water indicated. A chan...

Figure shows three temperature scales with the freezing and boiling point of water indicated. A change of 25 R⁰, 25 S⁰ and 25 U⁰ is denoted by x1x_1, x2x_2, x3x_3 respectively. Which of the following is correct:

A

x1>x2>x3x_1 > x_2 > x_3

B

x2<x1<x3x_2 < x_1 < x_3

C

x3>x2>x1x_3 > x_2 > x_1

D

can not be predicted

Answer

x3>x2>x1x_3 > x_2 > x_1

Explanation

Solution

The freezing and boiling points of water are 0°C and 100°C, respectively, giving a range of 100°C.

For scale R: Freezing point = -80°R, Boiling point = 120°R. Range = 120(80)=200120 - (-80) = 200 R⁰. Conversion: 200 R⁰ = 100°C     \implies 1 R⁰ = 0.5°C. x1=25x_1 = 25 R⁰ =25×0.5= 25 \times 0.5 °C =12.5= 12.5 °C.

For scale S: Freezing point = 50°S, Boiling point = 225°S. Range = 22550=175225 - 50 = 175 S⁰. Conversion: 175 S⁰ = 100°C     \implies 1 S⁰ = 100175\frac{100}{175} °C =47= \frac{4}{7} °C. x2=25x_2 = 25 S⁰ =25×47= 25 \times \frac{4}{7} °C =1007= \frac{100}{7} °C 14.286\approx 14.286 °C.

For scale U: Freezing point = 30°U, Boiling point = 120°U. Range = 12030=90120 - 30 = 90 U⁰. Conversion: 90 U⁰ = 100°C     \implies 1 U⁰ = 10090\frac{100}{90} °C =109= \frac{10}{9} °C. x3=25x_3 = 25 U⁰ =25×109= 25 \times \frac{10}{9} °C =2509= \frac{250}{9} °C 27.778\approx 27.778 °C.

Comparing the values: x1=12.5x_1 = 12.5 °C x214.286x_2 \approx 14.286 °C x327.778x_3 \approx 27.778 °C

Therefore, x1<x2<x3x_1 < x_2 < x_3, which can also be written as x3>x2>x1x_3 > x_2 > x_1.