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UPSC Civil Service Exam Questions - Engineering

Q1:

A bar 4 cm in diameter is subjected to an axial load of 41. The extension of the bar over a gauge length of 20 cm is 0.03 cm. The decrease in diameter is 0.0018 cm. The Poisson's ratio is

A 0.25

B 0.30

C 0.33

D 0.35

ANS:B - 0.30

μ = - εt / εl= -(-0.00045) /0.0015= 0.3 ---- (1)

Where,

μ = Poisson's ratio.

εt = transverse strain.

εl = longitudinal or axial strain.

Strain can be expressed as,

εt = dl / L = 0.0018/4=0.00045 , εl=0.0015.
Where,

dl = change in length (m, ft).

L = initial length (m, ft).