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A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will be
- a)K
- b)1.25 K
- c)1.56 K
- d)1.95 K
Correct answer is option 'C'. Can you explain this answer?
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A length of 10 mm diameter steel wire is coiled to a close helical sp...
Stiffness of the spring is given by
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D is the mean diameter of the spring coil, d is the spring wire diameter, n is the number of active coils, and G is the modulus of rigidity for the spring material.
Here d’ = 2d; D’ = 2D
Here G and d are the same.
k2 = 1.56 k1 = 1.56 K
Most Upvoted Answer
A length of 10 mm diameter steel wire is coiled to a close helical sp...
Given Data:
- Diameter of steel wire = 10 mm
- Coils in first scenario: 8 coils of 75 mm mean diameter
- Coils in second scenario: 10 coils of 60 mm mean diameter
Calculating Stiffness (K) for the First Scenario:
1. Radius of the wire = 10/2 = 5 mm = 0.005 m
2. Mean diameter of the first scenario = 75 mm = 0.075 m
3. Length of wire per coil = π * 0.005 * 8 = 0.1257 m
4. Total length of wire for 8 coils = 8 * 0.1257 = 1.0056 m
5. Stiffness (K) for the first scenario = (G * d^4) / (64 * D^3 * n * Dm)
where G is the shear modulus, d is the wire diameter, D is the mean diameter, n is the number of coils, and Dm is the mean diameter
6. Substitute the values to find K for the first scenario.
Calculating Stiffness (K) for the Second Scenario:
1. Mean diameter of the second scenario = 60 mm = 0.06 m
2. Length of wire per coil = π * 0.005 * 10 = 0.157 m
3. Total length of wire for 10 coils = 10 * 0.157 = 1.57 m
4. Stiffness (K) for the second scenario = (G * d^4) / (64 * D^3 * n * Dm)
5. Substitute the values to find K for the second scenario.
Comparing the Two Scenarios:
- Calculate the ratio of K for the second scenario to K for the first scenario
- K(second scenario) / K(first scenario) = 1.56
- Therefore, the spring stiffness for the second scenario will be 1.56 times the stiffness of the first scenario, which is represented as option 'C'.
- Diameter of steel wire = 10 mm
- Coils in first scenario: 8 coils of 75 mm mean diameter
- Coils in second scenario: 10 coils of 60 mm mean diameter
Calculating Stiffness (K) for the First Scenario:
1. Radius of the wire = 10/2 = 5 mm = 0.005 m
2. Mean diameter of the first scenario = 75 mm = 0.075 m
3. Length of wire per coil = π * 0.005 * 8 = 0.1257 m
4. Total length of wire for 8 coils = 8 * 0.1257 = 1.0056 m
5. Stiffness (K) for the first scenario = (G * d^4) / (64 * D^3 * n * Dm)
where G is the shear modulus, d is the wire diameter, D is the mean diameter, n is the number of coils, and Dm is the mean diameter
6. Substitute the values to find K for the first scenario.
Calculating Stiffness (K) for the Second Scenario:
1. Mean diameter of the second scenario = 60 mm = 0.06 m
2. Length of wire per coil = π * 0.005 * 10 = 0.157 m
3. Total length of wire for 10 coils = 10 * 0.157 = 1.57 m
4. Stiffness (K) for the second scenario = (G * d^4) / (64 * D^3 * n * Dm)
5. Substitute the values to find K for the second scenario.
Comparing the Two Scenarios:
- Calculate the ratio of K for the second scenario to K for the first scenario
- K(second scenario) / K(first scenario) = 1.56
- Therefore, the spring stiffness for the second scenario will be 1.56 times the stiffness of the first scenario, which is represented as option 'C'.
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A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. Can you explain this answer?
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A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. 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 A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. Can you explain this answer?.
A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. 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 A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. Can you explain this answer?.
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A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A length of 10 mm diameter steel wire is coiled to a close helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness K. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will bea)Kb)1.25 Kc)1.56 Kd)1.95 KCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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