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UPSC Civil Service Exam Questions - Engineering

Q1:

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

A twice

B four times

C eight times

D sixteen times

ANS:D - sixteen times

K = p/(δ) in spring.
The formula is k=Gd^4/(64R^3n).
So, (2d)^4= 16d^4.