Given, Decrement in the length =1% Poisson's ratio for material of the rod, σ=0.2 As we know that, Volume, v=πr2l,[ where ,l is the length of the rod]
∆V
V
=
2∆r
r
+
∆l
l
...(i) since, Poisson's ratio, σ=
−∆D/D
∆L/L
=
−∆r/r
∆L/L
So, Eq. (i) can be written as,
∆V
V
=
∆l
l
(1−2σ)
∆V
V
×100=(
∆l
l
×100)×(1−2σ)
=−1×[1−2×(0.2)] =−1×[1−0.4] =−0.6% Here, negative sign shows the decrement in the volume.