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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?
- a)10%
- b)5%
- c)1%
- d)0.5%
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A bar of length L tapers uniformly from diameter 1.1 D at one end to ...
Actual elongation,
Elongation computed using mean diameter,
Most Upvoted Answer
A bar of length L tapers uniformly from diameter 1.1 D at one end to ...
Let's assume the length of the bar is L and the diameter at one end is 1.1D, and at the other end is 0.9D.
To calculate the elongation due to axial pull, we use the mean diameter D. Let's denote the original length of the bar as Lo and the final length after elongation as Lf. The elongation (∆L) can be calculated using the formula:
∆L = (Lf - Lo)
To find the error in the computed elongation, we need to compare it with the actual elongation (∆L_actual). The actual elongation can be calculated by considering the varying diameter along the length of the bar.
Let's assume that the diameter varies linearly along the length of the bar. This means that the diameter at any point x along the length of the bar can be given by:
d = (1.1 - 0.2x)D
where x varies from 0 (one end of the bar) to L (the other end of the bar).
Now, let's calculate the actual elongation (∆L_actual) using the varying diameter:
∆L_actual = ∫[0,L] (1.1 - 0.2x)D dx
∆L_actual = D * [1.1x - 0.1x^2] evaluated from x = 0 to x = L
∆L_actual = D * [1.1L - 0.1L^2]
Now, let's calculate the elongation (∆L) using the mean diameter D:
∆L = D * (Lf - Lo)
Since the bar tapers uniformly, we can assume that the elongation is also uniform. Therefore, we can write:
∆L = D * (L - Lo)
To find the error in the computed elongation, we can calculate the percentage error as:
Error = [(∆L_actual - ∆L) / ∆L_actual] * 100%
Plugging in the values:
Error = [(D * [1.1L - 0.1L^2] - D * (L - Lo)) / (D * [1.1L - 0.1L^2])] * 100%
Simplifying the equation:
Error = [(1.1L - 0.1L^2 - L + Lo) / (1.1L - 0.1L^2)] * 100%
Since we are interested in the approximate error, let's assume that L is much larger than Lo and L^2 is much smaller compared to L. This allows us to simplify the equation further:
Error ≈ [(1.1L - L) / (1.1L)] * 100%
Error ≈ [(0.1L) / (1.1L)] * 100%
Error ≈ 0.1/1.1 * 100%
Error ≈ 9.09%
Therefore, the approximate error in the computed elongation is 9.09%, which is closest to option C, 1%.
To calculate the elongation due to axial pull, we use the mean diameter D. Let's denote the original length of the bar as Lo and the final length after elongation as Lf. The elongation (∆L) can be calculated using the formula:
∆L = (Lf - Lo)
To find the error in the computed elongation, we need to compare it with the actual elongation (∆L_actual). The actual elongation can be calculated by considering the varying diameter along the length of the bar.
Let's assume that the diameter varies linearly along the length of the bar. This means that the diameter at any point x along the length of the bar can be given by:
d = (1.1 - 0.2x)D
where x varies from 0 (one end of the bar) to L (the other end of the bar).
Now, let's calculate the actual elongation (∆L_actual) using the varying diameter:
∆L_actual = ∫[0,L] (1.1 - 0.2x)D dx
∆L_actual = D * [1.1x - 0.1x^2] evaluated from x = 0 to x = L
∆L_actual = D * [1.1L - 0.1L^2]
Now, let's calculate the elongation (∆L) using the mean diameter D:
∆L = D * (Lf - Lo)
Since the bar tapers uniformly, we can assume that the elongation is also uniform. Therefore, we can write:
∆L = D * (L - Lo)
To find the error in the computed elongation, we can calculate the percentage error as:
Error = [(∆L_actual - ∆L) / ∆L_actual] * 100%
Plugging in the values:
Error = [(D * [1.1L - 0.1L^2] - D * (L - Lo)) / (D * [1.1L - 0.1L^2])] * 100%
Simplifying the equation:
Error = [(1.1L - 0.1L^2 - L + Lo) / (1.1L - 0.1L^2)] * 100%
Since we are interested in the approximate error, let's assume that L is much larger than Lo and L^2 is much smaller compared to L. This allows us to simplify the equation further:
Error ≈ [(1.1L - L) / (1.1L)] * 100%
Error ≈ [(0.1L) / (1.1L)] * 100%
Error ≈ 0.1/1.1 * 100%
Error ≈ 9.09%
Therefore, the approximate error in the computed elongation is 9.09%, which is closest to option C, 1%.
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A bar of length L tapers uniformly from diameter 1.1 D at one end to ...
Actual elongation,
Elongation computed using mean diameter,
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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a) 10%b) 5%c) 1%d) 0.5%Correct answer is option 'C'. Can you explain this answer?
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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a) 10%b) 5%c) 1%d) 0.5%Correct 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 bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a) 10%b) 5%c) 1%d) 0.5%Correct 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 bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a) 10%b) 5%c) 1%d) 0.5%Correct answer is option 'C'. Can you explain this answer?.
A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a) 10%b) 5%c) 1%d) 0.5%Correct 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 bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a) 10%b) 5%c) 1%d) 0.5%Correct 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 bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a) 10%b) 5%c) 1%d) 0.5%Correct answer is option 'C'. Can you explain this answer?.
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