Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 2 (Q.No. 19)
19.
In a close-coiled helical spring subjected to an axial load, other quantities remaining the same, if the wire diameter is doubled, then the stiffness of the spring when compared to the original one, will become
Discussion:
4 comments Page 1 of 1.
Bapuji said:
7 years ago
Please, anyone explain the answer.
Mnb said:
6 years ago
Stiffness = Gd^4/64R^3n.
2^4 = 16.
2^4 = 16.
Shubham said:
5 years ago
It should be twice.
Ajay Sharma said:
4 years ago
K = p/(δ) in spring.
The formula is k=Gd^4/(64R^3n).
So, (2d)^4= 16d^4.
The formula is k=Gd^4/(64R^3n).
So, (2d)^4= 16d^4.
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