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Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B'. The stiffness of spring 'A' will be .....that of spring 'A'.
- a)one-sixteenth
- b)one- eighth
- c)one-fourth
- d)one-half
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Two closely coiled helical springs 'A' and 'B' are equal in all respec...
k = GD^4 / 8D^3 N.
Here d is small dia mean wire dia, as Stiffness is directly proortinal to d^4. so the answer is 1/16 of another spring.
Most Upvoted Answer
Two closely coiled helical springs 'A' and 'B' are equal in all respec...
Answer:
The stiffness of a spring is inversely proportional to the diameter of the wire used to make the spring. Therefore, the stiffness of spring B will be greater than that of spring A.
Explanation:
Stiffness of a spring is defined as the ratio of the force applied to the spring to the resulting deformation or displacement of the spring. The stiffness of a spring depends on various factors such as the material used to make the spring, the diameter of the wire used, the number of coils, and the length of the spring.
In this case, two closely coiled helical springs A and B are equal in all respects except for the diameter of wire used. The diameter of wire used in spring A is double that of spring B. Let us assume that the stiffness of spring A is kA and the stiffness of spring B is kB.
Now, as per the formula for stiffness, k ∝ 1/d^4, where d is the diameter of the wire used. Therefore, we can write:
kA ∝ 1/(2d)^4 => kA ∝ 1/16d^4
kB ∝ 1/d^4
Dividing the equations, we get:
kB/kA = (1/d^4)/(1/16d^4) = 16/1
Therefore, the stiffness of spring B is 16 times that of spring A.
Answer: (a) one-sixteenth.
The stiffness of a spring is inversely proportional to the diameter of the wire used to make the spring. Therefore, the stiffness of spring B will be greater than that of spring A.
Explanation:
Stiffness of a spring is defined as the ratio of the force applied to the spring to the resulting deformation or displacement of the spring. The stiffness of a spring depends on various factors such as the material used to make the spring, the diameter of the wire used, the number of coils, and the length of the spring.
In this case, two closely coiled helical springs A and B are equal in all respects except for the diameter of wire used. The diameter of wire used in spring A is double that of spring B. Let us assume that the stiffness of spring A is kA and the stiffness of spring B is kB.
Now, as per the formula for stiffness, k ∝ 1/d^4, where d is the diameter of the wire used. Therefore, we can write:
kA ∝ 1/(2d)^4 => kA ∝ 1/16d^4
kB ∝ 1/d^4
Dividing the equations, we get:
kB/kA = (1/d^4)/(1/16d^4) = 16/1
Therefore, the stiffness of spring B is 16 times that of spring A.
Answer: (a) one-sixteenth.
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Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B'. The stiffness of spring 'A' will be .....that of spring 'A'.a)one-sixteenthb)one- eighthc)one-fourthd)one-halfCorrect answer is option 'A'. Can you explain this answer?
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