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A closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring is
- a)50 N/cm
- b)250 N/cm
- c)100 N/cm
- d)500 N/cm
Correct answer is option 'C'. Can you explain this answer?
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A closely coiled spring of 20 cm mean diameter is having 25 coils of ...
Given data:
- Mean diameter of closely coiled spring = 20 cm
- Number of coils = 25
- Diameter of each coil = 2 cm
- Modulus of rigidity = 107 N/cm²
Calculating stiffness of the spring:
- First, we need to find the effective diameter of the spring using the given data.
- The mean diameter of the spring is given as 20 cm, and each coil has a diameter of 2 cm.
- Therefore, the effective diameter of the spring can be calculated as:
Effective diameter = Mean diameter + Diameter of one coil
= 20 cm + 2 cm
= 22 cm
- Now, we can calculate the radius of the spring:
Radius = Effective diameter / 2
= 22 cm / 2
= 11 cm
- The formula to calculate the stiffness of a spring is given by:
Stiffness = (Modulus of rigidity * (π * r^4)) / (8 * N * D^3)
where:
- r = radius of the spring
- N = number of coils
- D = diameter of the spring
- Substituting the values into the formula, we get:
Stiffness = (107 * (π * 11^4)) / (8 * 25 * 2^3)
≈ 100 N/cm
Therefore, the stiffness of the spring is 100 N/cm, which corresponds to option 'C'.
- Mean diameter of closely coiled spring = 20 cm
- Number of coils = 25
- Diameter of each coil = 2 cm
- Modulus of rigidity = 107 N/cm²
Calculating stiffness of the spring:
- First, we need to find the effective diameter of the spring using the given data.
- The mean diameter of the spring is given as 20 cm, and each coil has a diameter of 2 cm.
- Therefore, the effective diameter of the spring can be calculated as:
Effective diameter = Mean diameter + Diameter of one coil
= 20 cm + 2 cm
= 22 cm
- Now, we can calculate the radius of the spring:
Radius = Effective diameter / 2
= 22 cm / 2
= 11 cm
- The formula to calculate the stiffness of a spring is given by:
Stiffness = (Modulus of rigidity * (π * r^4)) / (8 * N * D^3)
where:
- r = radius of the spring
- N = number of coils
- D = diameter of the spring
- Substituting the values into the formula, we get:
Stiffness = (107 * (π * 11^4)) / (8 * 25 * 2^3)
≈ 100 N/cm
Therefore, the stiffness of the spring is 100 N/cm, which corresponds to option 'C'.
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A closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect answer is option 'C'. Can you explain this answer?
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A closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect 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 closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect 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 closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect answer is option 'C'. Can you explain this answer?.
A closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect 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 closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect 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 closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect answer is option 'C'. Can you explain this answer?.
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A closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A closely coiled spring of 20 cm mean diameter is having 25 coils of 2 cm diameter. Modulus of rigidity of the material is 107 N/cm2 . Stiffness of spring isa) 50 N/cmb) 250 N/cmc) 100 N/cmd) 500 N/cmCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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