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Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars is
- a)⅓
- b)½
- c)⅔
- d)3/2
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Two bars each of length I and 4 the same material are subjected to th...
For 1st bar,
Stress
Strain energy stored =
2nd bar, AB = F1
2nd bar, AB = F1
Stress energy = AB + BC
Most Upvoted Answer
Two bars each of length I and 4 the same material are subjected to th...
Explanation:
Given:
- Two bars of the same material
- Length of each bar = L
- Tensile force applied on both bars = P
- Diameter of the first bar = 2d
- Diameter of the second bar = d for a length of 2d, and 2d for the remaining length
To find:
The ratio of strain energies of the two bars
Strain Energy:
Strain energy is the energy stored in a material when it is deformed under load. It can be calculated using the formula:
Strain Energy (U) = (1/2) * Stress * Strain * Volume
Where:
Stress = Force / Area
Strain = Change in length / Original length
Volume = Cross-sectional area * Length
Calculating Strain Energy for the first bar:
- Diameter of the first bar = 2d
- Radius (r) = (2d)/2 = d
- Cross-sectional area (A) = π * r^2 = π * d^2
- Volume (V) = A * L = π * d^2 * L
Now, let's calculate the strain energy (U1) for the first bar:
- Stress = P / A
- Strain = Change in length / Original length = L / L = 1
Substituting the values into the formula:
U1 = (1/2) * (P / A) * 1 * (π * d^2 * L)
Calculating Strain Energy for the second bar:
The second bar has two different diameters:
- Diameter of the first part = d
- Diameter of the second part = 2d
We need to calculate the strain energy separately for each part and then add them.
For the first part:
- Radius (r1) = d/2
- Cross-sectional area (A1) = π * (d/2)^2 = π * d^2 / 4
- Volume (V1) = A1 * L = (π * d^2 / 4) * L
Now, let's calculate the strain energy (U1) for the first part:
- Stress = P / A1
- Strain = Change in length / Original length = L / L = 1
Substituting the values into the formula:
U1 = (1/2) * (P / A1) * 1 * ((π * d^2 / 4) * L)
For the second part:
- Radius (r2) = 2d/2 = d
- Cross-sectional area (A2) = π * (d)^2 = π * d^2
- Volume (V2) = A2 * L = (π * d^2) * L
Now, let's calculate the strain energy (U2) for the second part:
- Stress = P / A2
- Strain = Change in length / Original length = L / L = 1
Substituting the values into the formula:
U2 = (1/2) * (P / A2) * 1 * ((π * d^2) * L)
Total strain energy for the second bar:
U2 = U1 + U2 = (1/2) *
Given:
- Two bars of the same material
- Length of each bar = L
- Tensile force applied on both bars = P
- Diameter of the first bar = 2d
- Diameter of the second bar = d for a length of 2d, and 2d for the remaining length
To find:
The ratio of strain energies of the two bars
Strain Energy:
Strain energy is the energy stored in a material when it is deformed under load. It can be calculated using the formula:
Strain Energy (U) = (1/2) * Stress * Strain * Volume
Where:
Stress = Force / Area
Strain = Change in length / Original length
Volume = Cross-sectional area * Length
Calculating Strain Energy for the first bar:
- Diameter of the first bar = 2d
- Radius (r) = (2d)/2 = d
- Cross-sectional area (A) = π * r^2 = π * d^2
- Volume (V) = A * L = π * d^2 * L
Now, let's calculate the strain energy (U1) for the first bar:
- Stress = P / A
- Strain = Change in length / Original length = L / L = 1
Substituting the values into the formula:
U1 = (1/2) * (P / A) * 1 * (π * d^2 * L)
Calculating Strain Energy for the second bar:
The second bar has two different diameters:
- Diameter of the first part = d
- Diameter of the second part = 2d
We need to calculate the strain energy separately for each part and then add them.
For the first part:
- Radius (r1) = d/2
- Cross-sectional area (A1) = π * (d/2)^2 = π * d^2 / 4
- Volume (V1) = A1 * L = (π * d^2 / 4) * L
Now, let's calculate the strain energy (U1) for the first part:
- Stress = P / A1
- Strain = Change in length / Original length = L / L = 1
Substituting the values into the formula:
U1 = (1/2) * (P / A1) * 1 * ((π * d^2 / 4) * L)
For the second part:
- Radius (r2) = 2d/2 = d
- Cross-sectional area (A2) = π * (d)^2 = π * d^2
- Volume (V2) = A2 * L = (π * d^2) * L
Now, let's calculate the strain energy (U2) for the second part:
- Stress = P / A2
- Strain = Change in length / Original length = L / L = 1
Substituting the values into the formula:
U2 = (1/2) * (P / A2) * 1 * ((π * d^2) * L)
Total strain energy for the second bar:
U2 = U1 + U2 = (1/2) *
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Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer?
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Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer?.
Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer?.
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Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Two bars each of length I and 4 the same material are subjected to the same axial tensile force P. The first bar has a uniform diameter of 2d. The second bar has a diameter d for '2d' for the remaining length; the ratio of strain energies of the two bars isa)⅓b)½c)⅔d)3/2Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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