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A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then
1. E = 210 GPa and G = 77 GPa
2. E = 199 GPa and V = 0.25
3. E = 199 GPa and V = 0.30
4. E = 199 GPa and G = 80 GPa
Which of these values are correct?
- a)3 and 4
- b)2 and 4
- c)1and 3
- d)1 and 4
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN...
Given information:
- Diameter of the bar: 16 mm
- Elongation of the bar: 0.04%
- Tensile force applied: 16 kN
- Average decrease in diameter: 0.01%
Calculations:
1. Young's modulus (E) and Shear modulus (G) values are given as options. We need to calculate the Poisson's ratio (V) for each option and compare it with the given value.
2. Poisson's ratio (V) can be calculated using the formula:
V = (E - 2G) / (2E - 2G)
3. For option 1 (E = 210 GPa and G = 77 GPa), substituting the given values in the formula:
V = (210 - 2 * 77) / (2 * 210 - 2 * 77)
V ≈ 0.295
4. For option 2 (E = 199 GPa and V = 0.25), the given value of V can be directly used.
5. For option 3 (E = 199 GPa and V = 0.30), the given value of V can be directly used.
6. For option 4 (E = 199 GPa and G = 80 GPa), substituting the given values in the formula:
V = (199 - 2 * 80) / (2 * 199 - 2 * 80)
V ≈ 0.250
7. Now, we need to calculate the actual value of Poisson's ratio (V_actual) using the elongation and decrease in diameter.
8. The formula for elongation (ε) is given by:
ε = ΔL / L
where ΔL is the change in length and L is the original length.
9. The change in length can be calculated using the tensile force (F) and the cross-sectional area (A) of the bar:
ΔL = F * L / (E * A)
where A = π * (d^2) / 4
10. The decrease in diameter (Δd) can be calculated as:
Δd = d * V_actual
11. Rearranging the above formulas, we get:
V_actual = Δd / d
12. Substituting the given values, we can calculate V_actual for each option.
Results:
- For option 1, V_actual ≈ 0.01 / 16 ≈ 0.000625
- For option 2, V_actual ≈ 0.01 / 16 ≈ 0.000625
- For option 3, V_actual ≈ 0.01 / 16 ≈ 0.000625
- For option 4, V_actual ≈ 0.01 / 16 ≈ 0.000625
Conclusion:
- Comparing the calculated values of V_actual with the given values of V for each option, we can see that options 2 and 4 have the correct values. Thus, the correct options are b) 2 and 4.
- Diameter of the bar: 16 mm
- Elongation of the bar: 0.04%
- Tensile force applied: 16 kN
- Average decrease in diameter: 0.01%
Calculations:
1. Young's modulus (E) and Shear modulus (G) values are given as options. We need to calculate the Poisson's ratio (V) for each option and compare it with the given value.
2. Poisson's ratio (V) can be calculated using the formula:
V = (E - 2G) / (2E - 2G)
3. For option 1 (E = 210 GPa and G = 77 GPa), substituting the given values in the formula:
V = (210 - 2 * 77) / (2 * 210 - 2 * 77)
V ≈ 0.295
4. For option 2 (E = 199 GPa and V = 0.25), the given value of V can be directly used.
5. For option 3 (E = 199 GPa and V = 0.30), the given value of V can be directly used.
6. For option 4 (E = 199 GPa and G = 80 GPa), substituting the given values in the formula:
V = (199 - 2 * 80) / (2 * 199 - 2 * 80)
V ≈ 0.250
7. Now, we need to calculate the actual value of Poisson's ratio (V_actual) using the elongation and decrease in diameter.
8. The formula for elongation (ε) is given by:
ε = ΔL / L
where ΔL is the change in length and L is the original length.
9. The change in length can be calculated using the tensile force (F) and the cross-sectional area (A) of the bar:
ΔL = F * L / (E * A)
where A = π * (d^2) / 4
10. The decrease in diameter (Δd) can be calculated as:
Δd = d * V_actual
11. Rearranging the above formulas, we get:
V_actual = Δd / d
12. Substituting the given values, we can calculate V_actual for each option.
Results:
- For option 1, V_actual ≈ 0.01 / 16 ≈ 0.000625
- For option 2, V_actual ≈ 0.01 / 16 ≈ 0.000625
- For option 3, V_actual ≈ 0.01 / 16 ≈ 0.000625
- For option 4, V_actual ≈ 0.01 / 16 ≈ 0.000625
Conclusion:
- Comparing the calculated values of V_actual with the given values of V for each option, we can see that options 2 and 4 have the correct values. Thus, the correct options are b) 2 and 4.
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Community Answer
A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN...
Load (P) = 16kN
Diameter (d) = 16mm
% elongation = 0.04%
= 198.94 x 103MPa
= 199 GPa
= 79.6
= 80 GPa
Statement 2 and 4 are correct.
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Question Description
A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer?.
A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer?.
Solutions for A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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Here you can find the meaning of A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 16 mm diameter bar elongates by 0.04% under a tensile force of 16kN. The average decrease in diameter is found to be 0.01%. Then1. E = 210 GPa and G = 77 GPa2. E = 199 GPa and V = 0.253. E = 199 GPa and V = 0.304. E = 199 GPa and G = 80 GPaWhich of these values are correct?a) 3 and 4b) 2 and 4c) 1and 3d) 1 and 4Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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