Thermometer A and B have ice point marked at 15 and 25 and steam points a 75 and 125 respectively. When thermo-meter A measures the temperature of a bath as 60, the reading of B for the same bath is

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Q: 64 (NDA-II/2011)
Thermometer A and B have ice point marked at 15° and 25° and steam points a 75° and 125° respectively. When thermo-meter A measures the temperature of a bath as 60°, the reading of B for the same bath is

question_subject: 

Maths

question_exam: 

NDA-II

stats: 

0,4,6,1,1,4,4

keywords: 

{'temperature': [0, 1, 1, 7], 'ice point': [0, 0, 0, 1], 'steam points': [0, 0, 0, 1], 'thermo': [0, 0, 0, 1], 'meter': [0, 0, 0, 2], 'same bath': [0, 0, 0, 1], 'bath': [0, 0, 0, 1]}

Thermometer A and B have different ice and steam points. Thermometer A has an ice point marked at 15° and a steam point marked at 75°, while thermometer B has an ice point marked at 25° and a steam point marked at 125°.

To determine the reading of B for the same bath when thermometer A measures the temperature as 60°, we can use the concept of linear interpolation.

First, we calculate the difference between the ice point and steam point for both thermometers. For thermometer A, the difference is 75° - 15° = 60°. For thermometer B, the difference is 125° - 25° = 100°.

Then, we calculate the proportional relationship between the temperature difference and the actual measured temperature for thermometer A. In thermometer A, the measured temperature of 60° represents 60° / 60° = 1 proportion of the temperature difference.

Finally, we apply the proportional relationship to the temperature difference of thermometer B. Since the measured temperature of 60° on thermometer A represents 1 proportion of the temperature difference, it would also represent the same proportion on thermometer B. Thus, the reading of B for the same bath is 1 proportion of

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