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Maulik IyerEduRev Computer Science Engineering (CSE) |
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Maulik Iyer
EduRev Computer Science Engineering (CSE)
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Discussed Questions
Consider the below shown loading conditions, which one will fail first according to principal strain energy theory [μ = 0.3]I.
II.
- a) I will fail first
- b) II will fail first
- c) Both I and II will fail at a time
- d) Insufficient data to decide
Correct answer is option 'B'. Can you explain this answer?
Consider the below shown loading conditions, which one will fail first according to principal strain energy theory [μ = 0.3]
I.
II.
a)
I will fail first
b)
II will fail first
c)
Both I and II will fail at a time
d)
Insufficient data to decide
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Cstoppers Instructors answered • 3 weeks ago |
Σx = σ = σ1 similarly, σ2 = 2σ
σ3 = 3σ
Syt = 
Syt )I =
= 
= 
= √7.4 σ2
∵ [σ1 = 2σ, σ2 = σ, σ3 = 4σ]
Syt )II =
= 
= 
=√12.6σ2
Syt )II > Syt )I
A project consist of activities with the following data
The overhead cost per day is given as Rs.100. The minimum duration of the project with which minimum cost is associated ___________- a) 18
- b) 18
Correct answer is between ' 18, 18'. Can you explain this answer?
A project consist of activities with the following data
The overhead cost per day is given as Rs.100. The minimum duration of the project with which minimum cost is associated ___________
a)
18
b)
18
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Cstoppers Instructors answered • 3 weeks ago |
Paths:
A‐ D = 8 + 10 = 18
B‐ C‐ D = 5 + 5 + 10 = 19 (critical path)
(i) Total cost = direct cost + Indirect cost
Direct cost = 100 + 200 + 150 + 300 = 750 TC = 750 + 100 × 20
= Rs. 2750
(ii) crash activity ‘B’ as it is having minimum cost slope by ‘1’ days TC
= [750 + (40 × 1) + 100 × 19
= Rs. 2690
(iii) crash activity ‘D’ by 1 day TC = (750 + 40 × 1 + 100 × 1) + 100 × 18
= 2690
No further possibility to crash minimum cost
= 2690 Duration
= 18 days
A project consist of following activities
Minimum time required to complete the project in (days) is __________- a) 23
- b) 23
Correct answer is between ' 23, 23'. Can you explain this answer?
A project consist of following activities
Minimum time required to complete the project in (days) is __________
a)
23
b)
23
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Cstoppers Instructors answered • 3 weeks ago |
A‐ D‐ G = 4 + 5 + 8 = 17
A‐C‐E‐G = 4 + 6 + 5 + 8 = 23
B‐ E‐ G = 6 + 5 + 8 = 19
B‐ F = 6 + 6 = 12
A‐ C‐ F = 4 + 6 + 6 = 16
For the project given below, the number of critical paths are
- a) 1
- b) 2
- c) 3
- d) 4
Correct answer is option 'B'. Can you explain this answer?
For the project given below, the number of critical paths are
a)
1
b)
2
c)
3
d)
4
|
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Cstoppers Instructors answered • 3 weeks ago |
Paths:
1‐ 3‐ 5‐ 6 = 6 + 4 + 3 = 13∗
1‐ 2‐ 5‐ 6 = 8 + 2 + 3 = 13∗
1‐ 4‐ 6 = 7 + 5 = 12
Find the completion time of the project
(A)36(B)36
Correct answer is between ','. Can you explain this answer?
Find the completion time of the project
(A)36
(B)36
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Cstoppers Instructors answered • 3 weeks ago |
Paths:
A‐ C‐ G = 4 + 2 + 7 = 13
A‐D‐E‐G = 4 + 5 + 6 + 7 = 22
A‐ D‐ F‐ H‐ I = 4 + 5 + 6 + 10 + 11 = 36∗
B‐D‐E‐G = 3 + 5 + 6 + 7 = 21
B‐ D‐ F‐ H‐ I = 3 + 5 + 6 + 10 + 11 = 35
Completion time = 36 weeks
For the figure shown below, the permissible tensile and shear stresses for the weld material and plates are N mm and N mm . If the depth of weld is 15 mm, the length of each parallel fillet weld is
- a) 67 mm
- b) 103 mm
- c) 56 mm
- d) 135 mm
Correct answer is option 'A'. Can you explain this answer?
For the figure shown below, the permissible tensile and shear stresses for the weld material and plates are N mm and N mm . If the depth of weld is 15 mm, the length of each parallel fillet weld is
a)
67 mm
b)
103 mm
c)
56 mm
d)
135 mm
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Cstoppers Instructors answered • 4 weeks ago |
Strength of transverse fillet weld P1
P1 = 0.707 hl
t
t= 0.707 x (15)(100) x 80 = 84840 N
Strength of parallel fillet weld P2
P2 = 0.707 x 2 x hlτ
= 0.707 x 2 x 15 x l x 60 = 1272.6 l
Now, P1 + P2 = 170 x 1000
84840 + 1272.6 l = 170000
l = 66.98 
mm
mm Televisions are produced in batches. The production rate is 80 units/day. The demand during production period is 50 units/day. The set-up cost of manufacturing process is 3000 per set-up and holding cost is Rs. 15/unit/year. What is the economic batch size?- a) 5578
- b) 5582
Correct answer is between ' 5578, 5582'. Can you explain this answer?
Televisions are produced in batches. The production rate is 80 units/day. The demand during production period is 50 units/day. The set-up cost of manufacturing process is 3000 per set-up and holding cost is Rs. 15/unit/year. What is the economic batch size?
a)
5578
b)
5582
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Cstoppers Instructors answered • 4 weeks ago |
= Rs. 4412
The master production schedule to manufacture a product is given below:
If the company follows Wilson’s model to place an order. The EOQ is given as 300, the stock-on-hand is 100, lead time is 1 month. Find the minimum number of orders required to fulfill the demand arise up to month-8- a) 3
- b) 4
- c) 5
- d) 6
Correct answer is option 'B'. Can you explain this answer?
The master production schedule to manufacture a product is given below:
If the company follows Wilson’s model to place an order. The EOQ is given as 300, the stock-on-hand is 100, lead time is 1 month. Find the minimum number of orders required to fulfill the demand arise up to month-8
a)
3
b)
4
c)
5
d)
6
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Cstoppers Instructors answered • 4 weeks ago |
Minimum number of orders required is 4
Given that the direct fixed cost is Rs.4000. Find the optimal completion time of the project and the cost associated with it, if the overhead cost is Rs.260/day.
- a) 30,12310
- b) 30,12330
- c) 31,12310
- d) 32,12320
Correct answer is option 'C'. Can you explain this answer?
Given that the direct fixed cost is Rs.4000. Find the optimal completion time of the project and the cost associated with it, if the overhead cost is Rs.260/day.
a)
30,12310
b)
30,12330
c)
31,12310
d)
32,12320
|
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Cstoppers Instructors answered • 4 weeks ago |
Paths are:
A‐ B‐ E‐ F‐ G‐ I = 3 + 3 + 5 + 6 + 4 + 10 = 31 30
A‐ B‐ E‐ F‐ H = 3 + 3 + 5 + 6 + 10 = 27 26
A‐ C‐ F‐ G‐ I = 3 + 7 + 6 + 4 + 10 = 30
A‐C‐F‐H = 3 + 7 + 6 + 13 = 29
A‐D‐F‐G‐ I = 3 + 9 + 6 + 4 + 10 = 32∗ 31 30
A‐D‐F‐H = 3 + 9 + 6 + 13 = 31 30 29
(i) TC = 4000 + 260 × 32 = 12,320
(ii) Crash activity ‘D’ by 1 day TC = (4000 + 250 × 1) + 260 × 31
= 12,310
(iii) Crash B, D each by 1 day TC = (4000 + 250 × 1 + 30 × 1 + 250× 1) + 260 × 30
= 12,330
Optimal time = 31
Cost = 12310
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