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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...
Given Data:
- Diameter of brass rod: 20mm
- Tensile load applied: 40 kN
- Extension measured: 254 divisions in 200mm
- Each division on the extension meter: 0.001mm
To find:
The elastic modulus of brass.
Formula used:
The formula to calculate the elastic modulus (E) is given by:
E = (F * L) / (A * ΔL)
Where:
E = Elastic modulus
F = Applied force
L = Original length of the rod
A = Cross-sectional area of the rod
ΔL = Change in length of the rod
Calculation:
1. Convert the diameter of the rod to radius:
Radius (r) = Diameter / 2 = 20mm / 2 = 10mm = 10 * 0.001m = 0.01m
2. Calculate the original length of the rod:
Original length (L) = 200mm = 200 * 0.001m = 0.2m
3. Calculate the cross-sectional area of the rod:
Cross-sectional area (A) = π * r^2 = 3.1415 * (0.01m)^2 = 3.1415 * 0.0001m^2 = 0.00031415m^2
4. Convert the applied force to Newtons:
Force (F) = 40 kN = 40 * 1000N = 40000N
5. Calculate the change in length of the rod:
Change in length (ΔL) = (254 divisions * 0.001mm/division) = 0.254mm = 0.254 * 0.001m = 0.000254m
6. Substitute the values in the formula to calculate the elastic modulus:
E = (F * L) / (A * ΔL)
E = (40000N * 0.2m) / (0.00031415m^2 * 0.000254m)
E = 1.0025 x 10^5 N/mm^2
Therefore, the elastic modulus of brass is approximately 1.0025 x 10^5 N/mm^2, which corresponds to option A.
- Diameter of brass rod: 20mm
- Tensile load applied: 40 kN
- Extension measured: 254 divisions in 200mm
- Each division on the extension meter: 0.001mm
To find:
The elastic modulus of brass.
Formula used:
The formula to calculate the elastic modulus (E) is given by:
E = (F * L) / (A * ΔL)
Where:
E = Elastic modulus
F = Applied force
L = Original length of the rod
A = Cross-sectional area of the rod
ΔL = Change in length of the rod
Calculation:
1. Convert the diameter of the rod to radius:
Radius (r) = Diameter / 2 = 20mm / 2 = 10mm = 10 * 0.001m = 0.01m
2. Calculate the original length of the rod:
Original length (L) = 200mm = 200 * 0.001m = 0.2m
3. Calculate the cross-sectional area of the rod:
Cross-sectional area (A) = π * r^2 = 3.1415 * (0.01m)^2 = 3.1415 * 0.0001m^2 = 0.00031415m^2
4. Convert the applied force to Newtons:
Force (F) = 40 kN = 40 * 1000N = 40000N
5. Calculate the change in length of the rod:
Change in length (ΔL) = (254 divisions * 0.001mm/division) = 0.254mm = 0.254 * 0.001m = 0.000254m
6. Substitute the values in the formula to calculate the elastic modulus:
E = (F * L) / (A * ΔL)
E = (40000N * 0.2m) / (0.00031415m^2 * 0.000254m)
E = 1.0025 x 10^5 N/mm^2
Therefore, the elastic modulus of brass is approximately 1.0025 x 10^5 N/mm^2, which corresponds to 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 N/mm2Correct 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 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 N/mm2Correct 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 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 N/mm2Correct 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 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 N/mm2Correct 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 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 N/mm2Correct 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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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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 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 eachdivision is equal to 0.001mm, theelastic modulus of brass is :-a)1.0025 x105 N/mm2b)3.013 x105 N/mm2c)2.613 x105 N/mm2d)4.312 x105 N/mm2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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