question_answer 2)

Two absolute scales A and 5 have triple points of water defined to be 200 A and 350 B. What is the relation between \[{{\text{Y}}_{\text{steel}}}\text{=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}\text{N}{{\text{m}}^{\text{-2}}}\text{ and}\]and \[{{Y}_{alu\min ium}}=7\cdot 0\times \text{1}{{\text{0}}^{\text{10}}}\text{ N}{{\text{m}}^{\text{-2}}}\]?

Answer:

Given, triple point of water on scale A = 200 A Triple point of water on scale B = 350 B. As per question, 200 A = 350 B= 273\[{{\text{l}}_{\text{1}}}\text{=l;}{{\text{A}}_{\text{1}}}\text{=m}{{\text{m}}^{\text{2}}}\text{;}\]16 K or \[{{\text{Y}}_{\text{1}}}\text{=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}\text{N}{{\text{m}}^{\text{-2}}}\] and \[\text{B,}{{\text{l}}_{\text{2}}}\text{=l;}{{\text{A}}_{\text{2}}}\text{=m}{{\text{m}}^{\text{2}}}\text{;}\] If \[\frac{{{T}_{F}}'-32}{180}=\frac{{{T}_{K}}'-273\cdot 15}{100}\]represent the triple point of water on two scales A and B, then \[\frac{{{T}_{F}}'-{{T}_{F}}}{180}=\frac{{{T}_{K}}'-{{T}_{K}}}{100}\] or \[{{T}_{F}}'-{{T}_{F}}=\frac{180}{100}\left( {{T}_{K}}'-{{T}_{K}} \right)\] or \[{{T}_{K}}'-{{T}_{K}}=1K,\]