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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'.
Answer & Solution
Correct Answer:
Option
A
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
$$\left( \frac{\text{Deflection of Spring B}}{\text{Deflection of Spring A}} \right) = \left( \frac{\text{Mean Diameter of Spring A}}{\text{Mean Diameter of Spring B}} \right)^3$$
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
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Login(def1/def2) = (50/25)3
def(1/2)=2^3=8