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Brass rod 20mm in diameter is subjected to tensile load of 40 kN. Extension of rod measured with extension meter is found to be 254 divisions in 200mm. If each division is equal to 0.001mm, the elastic modulus of brass is :-
- a)1.0025 x 105 N/mm2
- b)3.013 x 105 N/mm2
- c)2.613 x 105 N/mm2
- d)4.312 x 105 N/mm2
Correct answer is option 'A'. Can you explain this answer?
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Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extens...
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Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extens...
Calculation of Elastic Modulus of Brass Rod
Given parameters:
Diameter of brass rod (d) = 20 mm
Tensile load (P) = 40 kN
Extension of rod (ΔL) = 254 divisions in 200 mm
One division (δL) = 0.001 mm
1. Calculation of cross-sectional area (A):
A = (π/4) x d^2
A = (π/4) x (20)^2
A = 314.16 mm^2
2. Calculation of strain (ε):
ε = ΔL/L
L = 200 mm (original length of rod)
ε = (254 x 0.001)/200
ε = 0.00127
3. Calculation of stress (σ):
σ = P/A
σ = 40,000 N/314.16 mm^2
σ = 127.3 N/mm^2
4. Calculation of elastic modulus (E):
σ = E x ε
E = σ/ε
E = 127.3 N/mm^2/0.00127
E = 100,250 N/mm^2
Therefore, the elastic modulus of brass is 1.0025 x 10^5 N/mm^2 (option A).
Given parameters:
Diameter of brass rod (d) = 20 mm
Tensile load (P) = 40 kN
Extension of rod (ΔL) = 254 divisions in 200 mm
One division (δL) = 0.001 mm
1. Calculation of cross-sectional area (A):
A = (π/4) x d^2
A = (π/4) x (20)^2
A = 314.16 mm^2
2. Calculation of strain (ε):
ε = ΔL/L
L = 200 mm (original length of rod)
ε = (254 x 0.001)/200
ε = 0.00127
3. Calculation of stress (σ):
σ = P/A
σ = 40,000 N/314.16 mm^2
σ = 127.3 N/mm^2
4. Calculation of elastic modulus (E):
σ = E x ε
E = σ/ε
E = 127.3 N/mm^2/0.00127
E = 100,250 N/mm^2
Therefore, the elastic modulus of brass is 1.0025 x 10^5 N/mm^2 (option A).
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Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer?
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Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer?.
Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer?.
Solutions for Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
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Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Brass rod 20mm in diameter issubjected to tensile load of 40 kN.Extension of rod measured withextension meter is found to be 254divisions in 200mm. If each division is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x 105 N/mm2b)3.013 x 105 N/mm2c)2.613 x 105 N/mm2d)4.312 x 105 N/mm2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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