SSC JE Mechanical Engineering 2017 Set 1 Paper

Question : 186 of 200

Marks: +1, -0

In a closed helical spring subjected to an axial load, other quantities remaining the same. if the wire diameter is doubled and mean radius of the coil is also doubled, then stiffness of spring when compared to original one will become______.

Twice
Four times
Eight times
Sixteen times

Validate

Solution:

Spring stiffness formula:

K={Gd^4}/{8D^3N}

where, d = wire diameter
D = Diameter of coil we get,

K_1={G2^4d^4}/{8×2^3D^3N}=2K

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