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A rod of length l tapers uniformly from a diameter D at one end to a diameter D/2 at the other end and is subjected to an axial load P. A second rod of length l and of uniform diameter D is subjected to the same axial load P. Both the rods are of same material with Young’s Modulus of elasticity E. The ratio of extension of the first rod to that of the second rod is
- a)4
- b)3
- c)2
- d)1
Correct answer is option 'C'. Can you explain this answer?
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A rod of length l tapers uniformly from a diameter D at one end to a ...
Extension of the tapered rod
Extension of the second rod of uniform diameter
This question is part of UPSC exam. View all Mechanical Engineering courses
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A rod of length l tapers uniformly from a diameter D at one end to a ...
To solve this problem, let's consider the elongation or extension of each rod separately.
For the first rod, which tapers uniformly from diameter D to D/2, we can assume that the tapering occurs linearly along the length of the rod. This means that the diameter changes uniformly from D to D/2, and therefore the cross-sectional area also changes uniformly.
Let's denote the original cross-sectional area of the rod as A₀ (corresponding to diameter D) and the cross-sectional area at any point along the length as A(x) (corresponding to diameter at that point).
Since the tapering is linear, we can express the cross-sectional area A(x) as a function of x, the distance from the end where the diameter is D. We can write:
A(x) = A₀ - (A₀ - Aₙ) * (x/l)
where Aₙ is the cross-sectional area at the other end of the rod (corresponding to diameter D/2) and l is the length of the rod.
Now, using Hooke's Law, we know that the stress (σ) in a rod is directly proportional to the strain (ε), which is the elongation divided by the original length. Mathematically, this can be written as:
σ = E * ε
where E is the Young's Modulus of elasticity.
Since the material and the axial load P are the same for both rods, the stress in both rods will be the same. Therefore, we can equate the stresses in the two rods:
E * ε₁ = E * ε₂
where ε₁ is the extension of the first rod and ε₂ is the extension of the second rod.
Now, we can calculate the extensions of the two rods using the formula for stress:
ε₁ = σ / E = P / (E * A₁)
ε₂ = σ / E = P / (E * A₀)
where A₁ is the cross-sectional area of the first rod, which can be calculated using the expression for A(x) as A(0) = A₀ and A(l) = Aₙ.
Now, substituting the expressions for A₁ and A₀ in terms of A₀ and Aₙ, we get:
ε₁ = P / (E * A₀)
ε₂ = P / (E * A₀)
Therefore, the ratio of the extension of the first rod to that of the second rod is:
ε₁ / ε₂ = (P / (E * A₀)) / (P / (E * A₀)) = 1
Hence, the correct answer is option 'd) 1'.
For the first rod, which tapers uniformly from diameter D to D/2, we can assume that the tapering occurs linearly along the length of the rod. This means that the diameter changes uniformly from D to D/2, and therefore the cross-sectional area also changes uniformly.
Let's denote the original cross-sectional area of the rod as A₀ (corresponding to diameter D) and the cross-sectional area at any point along the length as A(x) (corresponding to diameter at that point).
Since the tapering is linear, we can express the cross-sectional area A(x) as a function of x, the distance from the end where the diameter is D. We can write:
A(x) = A₀ - (A₀ - Aₙ) * (x/l)
where Aₙ is the cross-sectional area at the other end of the rod (corresponding to diameter D/2) and l is the length of the rod.
Now, using Hooke's Law, we know that the stress (σ) in a rod is directly proportional to the strain (ε), which is the elongation divided by the original length. Mathematically, this can be written as:
σ = E * ε
where E is the Young's Modulus of elasticity.
Since the material and the axial load P are the same for both rods, the stress in both rods will be the same. Therefore, we can equate the stresses in the two rods:
E * ε₁ = E * ε₂
where ε₁ is the extension of the first rod and ε₂ is the extension of the second rod.
Now, we can calculate the extensions of the two rods using the formula for stress:
ε₁ = σ / E = P / (E * A₁)
ε₂ = σ / E = P / (E * A₀)
where A₁ is the cross-sectional area of the first rod, which can be calculated using the expression for A(x) as A(0) = A₀ and A(l) = Aₙ.
Now, substituting the expressions for A₁ and A₀ in terms of A₀ and Aₙ, we get:
ε₁ = P / (E * A₀)
ε₂ = P / (E * A₀)
Therefore, the ratio of the extension of the first rod to that of the second rod is:
ε₁ / ε₂ = (P / (E * A₀)) / (P / (E * A₀)) = 1
Hence, the correct answer is option 'd) 1'.
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A rod of length l tapers uniformly from a diameter D at one end to a diameter D/2 at the other end and is subjected to an axial load P. A second rod of length l and of uniform diameter D is subjected to the same axial load P. Both the rods are of same material with Young’s Modulus of elasticity E. The ratio of extension of the first rod to that of the second rod isa) 4b) 3c) 2d) 1Correct answer is option 'C'. Can you explain this answer?
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A rod of length l tapers uniformly from a diameter D at one end to a diameter D/2 at the other end and is subjected to an axial load P. A second rod of length l and of uniform diameter D is subjected to the same axial load P. Both the rods are of same material with Young’s Modulus of elasticity E. The ratio of extension of the first rod to that of the second rod isa) 4b) 3c) 2d) 1Correct answer is option 'C'. 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 rod of length l tapers uniformly from a diameter D at one end to a diameter D/2 at the other end and is subjected to an axial load P. A second rod of length l and of uniform diameter D is subjected to the same axial load P. Both the rods are of same material with Young’s Modulus of elasticity E. The ratio of extension of the first rod to that of the second rod isa) 4b) 3c) 2d) 1Correct answer is option 'C'. 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 rod of length l tapers uniformly from a diameter D at one end to a diameter D/2 at the other end and is subjected to an axial load P. A second rod of length l and of uniform diameter D is subjected to the same axial load P. Both the rods are of same material with Young’s Modulus of elasticity E. The ratio of extension of the first rod to that of the second rod isa) 4b) 3c) 2d) 1Correct answer is option 'C'. Can you explain this answer?.
A rod of length l tapers uniformly from a diameter D at one end to a diameter D/2 at the other end and is subjected to an axial load P. A second rod of length l and of uniform diameter D is subjected to the same axial load P. Both the rods are of same material with Young’s Modulus of elasticity E. The ratio of extension of the first rod to that of the second rod isa) 4b) 3c) 2d) 1Correct answer is option 'C'. 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 rod of length l tapers uniformly from a diameter D at one end to a diameter D/2 at the other end and is subjected to an axial load P. A second rod of length l and of uniform diameter D is subjected to the same axial load P. Both the rods are of same material with Young’s Modulus of elasticity E. The ratio of extension of the first rod to that of the second rod isa) 4b) 3c) 2d) 1Correct answer is option 'C'. 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 rod of length l tapers uniformly from a diameter D at one end to a diameter D/2 at the other end and is subjected to an axial load P. A second rod of length l and of uniform diameter D is subjected to the same axial load P. Both the rods are of same material with Young’s Modulus of elasticity E. The ratio of extension of the first rod to that of the second rod isa) 4b) 3c) 2d) 1Correct answer is option 'C'. Can you explain this answer?.
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