consider two steel rods Aand B. B has the three times the and twice the length of A so the Young's modules of B will be what factor times Young's modules of A?
Answers
Answer:
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Young's modulus B is 1 times Young's modulus A or both Young’s modulus of A and B are the same.
Given: Two steel bars A and B. B has three times the cross-sectional area and twice the length of A.
To Find: The factor by which the Young's moduli of B are greater than the Young's moduli of A.
Solution:
- Young's modulus can be defined as the mechanical property of a material to resist compression or elongation in terms of its length.
- Young's modulus is a basic material property and does not depend on the length and cross-sectional area of a piece of material.
- The Young's modulus of a material depends only on the temperature and pressure of the material.
Coming to the question, steel bar B is said to have three times the cross-sectional area and twice the length of steel bar B.
Here we see that both bars are made of steel only. Because Young's modulus is a fundamental property of the material, it does not change with a change in dimensions.
So, Young's modulus of both bars A and B is the same.
Thus, Young's modulus B is 1 times higher than Young's modulus A or both Young’s modulus of A and B are the same.
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