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A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)
- a)91.8 MPa, 0.688
- b)71.8 MPa, 0.688
- c)91.8 MPa, 0.488
- d)71.8 MPa, 0.488
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A 10 mm diameter mild steel bar of length 1.50 metres is stressed by ...
!rea of the bar = π / 4× (102) = 78.54 mm2
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Let the maximum instantaneous stress be p N/mm2
∴ Maximum elongation = δl =
Equating the loss of potential energy to the strain energy stored by the member, we have
P(h + δl) =
∴ 2400 + 120
∴ p2 − 240 p/A=
∴ p2 −
∴ p2 − 3.056 p = 8148.714
∴ (p − 1.528)2 = 1848.714 + 2.334
∴ (p − 1.528)2 = 8151.049
∴ p − 1.528 = 90.283
p = 91.811 N/mm2
Alternatively, p =
But,
= 1.528 N/mm2
= 1.528 N/mm2= 1.5282 = 2.335
∴ p = 1.528 +
= 91.811 N/mm2
∴ Maximum elongation
= δl = 
= 0.6885 mm
Most Upvoted Answer
A 10 mm diameter mild steel bar of length 1.50 metres is stressed by ...
!rea of the bar = π / 4× (102) = 78.54 mm2
Let the maximum instantaneous stress be p N/mm2
∴ Maximum elongation = δl =
Equating the loss of potential energy to the strain energy stored by the member, we have
P(h + δl) =
∴ 2400 + 120
∴ p2 − 240 p/A=
∴ p2 −
∴ p2 − 3.056 p = 8148.714
∴ (p − 1.528)2 = 1848.714 + 2.334
∴ (p − 1.528)2 = 8151.049
∴ p − 1.528 = 90.283
p = 91.811 N/mm2
Alternatively, p =
But, = 1.528 N/mm2
= 1.528 N/mm2= 1.5282 = 2.335
∴ p = 1.528 +
= 91.811 N/mm2
∴ Maximum elongation
= δl =
= 0.6885 mm
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Community Answer
A 10 mm diameter mild steel bar of length 1.50 metres is stressed by ...
Maximum Instantaneous Stress Calculation:
- Weight of 120 N drops freely through 20 mm before stretching the bar.
- Weight = Force = mass x acceleration = 120 N
- Initial elongation = 20 mm = 0.02 m
- Stress = Force / Area = 120 / (π * (10/2)^2) = 91.8 MPa
Elongation Calculation:
- Using Hooke's Law: σ = E * ε
- ε = σ / E = 91.8 / 2 × 10^5 = 4.59 × 10^-4
- Elongation = ε * L = 4.59 × 10^-4 * 1.5 = 0.688 mm
Therefore, the maximum instantaneous stress in the bar is 91.8 MPa and the elongation produced in the bar is 0.688 mm. The correct answer is option 'A'.
- Weight of 120 N drops freely through 20 mm before stretching the bar.
- Weight = Force = mass x acceleration = 120 N
- Initial elongation = 20 mm = 0.02 m
- Stress = Force / Area = 120 / (π * (10/2)^2) = 91.8 MPa
Elongation Calculation:
- Using Hooke's Law: σ = E * ε
- ε = σ / E = 91.8 / 2 × 10^5 = 4.59 × 10^-4
- Elongation = ε * L = 4.59 × 10^-4 * 1.5 = 0.688 mm
Therefore, the maximum instantaneous stress in the bar is 91.8 MPa and the elongation produced in the bar is 0.688 mm. The correct answer is option 'A'.
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A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer?
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A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer?.
A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer?.
Solutions for A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm2)a) 91.8 MPa, 0.688b) 71.8 MPa, 0.688c) 91.8 MPa, 0.488d) 71.8 MPa, 0.488Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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