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An open coiled helical compression spring 'A' of mean diameter 50 mm is subjected to an axial load ‘W’. Another spring 'B' of mean diameter 25 mm is similar to spring 'A' in all respects. The deflection of spring 'B' will be __________ as compared to spring 'A'.

A. One-eighth

B. One-fourth

C. One-half

D. Double

Answer: Option A

Solution(By Examveda Team)

When comparing the deflection of two springs with different mean diameters, we use the formula:
$$\left( \frac{\text{Deflection of Spring B}}{\text{Deflection of Spring A}} \right) = \left( \frac{\text{Mean Diameter of Spring A}}{\text{Mean Diameter of Spring B}} \right)^3$$
Given that spring 'A' has a mean diameter of 50 mm and spring 'B' has a mean diameter of 25 mm:
$$\left( \frac{50}{25} \right)^3 = 2^3 = 8$$
This implies that the deflection of spring 'B' is one-eighth $$ \frac{1}{8} $$ of the deflection of spring 'A'.
Hence, the correct answer is:
Option A: One-eighth

This Question Belongs to Mechanical Engineering >> Machine Design

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Comments ( 2 )

  1. 18-333 Meghana
    18-333 Meghana :
    8 months ago

    def proportinal to cube(dia)

    (def1/def2) = (50/25)3
    def(1/2)=2^3=8

  2. Vinayak Bongirwar
    Vinayak Bongirwar :
    3 years ago

    deflection is directly proportional to cube of mean diameter

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