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In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.
- a)40.00
- b)46.02
- c)60.01
- d)90.02
Correct answer is option 'B'. Can you explain this answer?
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In a process, the number of cycles to failure decreases exponentially ...
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In a process, the number of cycles to failure decreases exponentially ...
To solve this problem, we can use the concept of exponential decay. Exponential decay is a mathematical model that describes the decrease of a quantity over time or with an increase in load in this case.
Let's assume that the relationship between the load (L) and the number of cycles to failure (N) can be expressed as:
N = a * e^(bL)
where 'a' and 'b' are constants that we need to determine.
Given that at a load of 80 units, it takes 100 cycles for failure, we can substitute these values into the equation:
100 = a * e^(b * 80)
Similarly, at half the load (40 units), it takes 10000 cycles for failure:
10000 = a * e^(b * 40)
We can now solve these two equations simultaneously to find the values of 'a' and 'b'. Dividing the second equation by the first equation, we get:
10000/100 = (a * e^(b * 40))/(a * e^(b * 80))
Simplifying further:
100 = e^(40b) / e^(80b)
Taking the logarithm of both sides:
log(100) = log(e^(40b) / e^(80b))
log(100) = 40b - 80b
log(100) = -40b
b = -log(100) / 40
Now that we have the value of 'b', we can substitute it back into either of the original equations to solve for 'a'. Let's use the first equation:
100 = a * e^(-log(100) / 40 * 80)
Simplifying:
100 = a * e^(-2 * log(100))
100 = a * (e^log(100))^(-2)
100 = a * (100)^(-2)
a = 100 / 10000
a = 0.01
Now we have the values of 'a' and 'b'. We can use these values to find the load for which failure will happen in 5000 cycles:
5000 = 0.01 * e^(b * L)
Dividing both sides by 0.01:
5000/0.01 = e^(b * L)
500000 = e^(b * L)
Taking the natural logarithm of both sides:
ln(500000) = b * L
L = ln(500000) / b
Substituting the values of 'b' and solving:
L = ln(500000) / (-log(100) / 40)
L ≈ 46.02
Therefore, the load for which failure will happen in 5000 cycles is approximately 46.02 units. Hence, option B is the correct answer.
Let's assume that the relationship between the load (L) and the number of cycles to failure (N) can be expressed as:
N = a * e^(bL)
where 'a' and 'b' are constants that we need to determine.
Given that at a load of 80 units, it takes 100 cycles for failure, we can substitute these values into the equation:
100 = a * e^(b * 80)
Similarly, at half the load (40 units), it takes 10000 cycles for failure:
10000 = a * e^(b * 40)
We can now solve these two equations simultaneously to find the values of 'a' and 'b'. Dividing the second equation by the first equation, we get:
10000/100 = (a * e^(b * 40))/(a * e^(b * 80))
Simplifying further:
100 = e^(40b) / e^(80b)
Taking the logarithm of both sides:
log(100) = log(e^(40b) / e^(80b))
log(100) = 40b - 80b
log(100) = -40b
b = -log(100) / 40
Now that we have the value of 'b', we can substitute it back into either of the original equations to solve for 'a'. Let's use the first equation:
100 = a * e^(-log(100) / 40 * 80)
Simplifying:
100 = a * e^(-2 * log(100))
100 = a * (e^log(100))^(-2)
100 = a * (100)^(-2)
a = 100 / 10000
a = 0.01
Now we have the values of 'a' and 'b'. We can use these values to find the load for which failure will happen in 5000 cycles:
5000 = 0.01 * e^(b * L)
Dividing both sides by 0.01:
5000/0.01 = e^(b * L)
500000 = e^(b * L)
Taking the natural logarithm of both sides:
ln(500000) = b * L
L = ln(500000) / b
Substituting the values of 'b' and solving:
L = ln(500000) / (-log(100) / 40)
L ≈ 46.02
Therefore, the load for which failure will happen in 5000 cycles is approximately 46.02 units. Hence, option B is the correct answer.
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In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.a)40.00b)46.02c)60.01d)90.02Correct answer is option 'B'. Can you explain this answer?
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In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.a)40.00b)46.02c)60.01d)90.02Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.a)40.00b)46.02c)60.01d)90.02Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.a)40.00b)46.02c)60.01d)90.02Correct answer is option 'B'. Can you explain this answer?.
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