Question

A wire of circular cross-section is stretched by
Longitudinal stress and its change
found to be 0.01%
If the Passion's ratio
for the material be 0.2 , what is the percentage
change in volume​

Answer 1

Answer:0.006%

Explanation:Let the length of the wire be l, its radius be r and volume be V.

So, V=πr2l

∴dV=πr2dl+2πlrdr[∵σ=−dr/rdl/l,dr=−rσdll]

=πr2dl−2πr2σdl

=πr2dl(l−2σ)

∴ Volume strain

=dVV=πr2dl(1−2σ)πr2l

=dll(1−2σ)[Here,dll=0.01%=0.0001]

=0.0001(1−2×0.2)=6×10−5

∴ Percentage change in volume

=dVV×100=6×10−5×100=0.006%