Question

If the volume of a wire remains constant
when subjected to tensile stress & the value
Poisson's ratio of material of the wire
of
is​

Answer 1

Explanation:

Now it is given that volume of the wire remains constant when subject to a tensile strength. Therefore differentiate both sides of the equation (1) and equate to zero as change in volume is zero (i.e. volume of the wire remains constant). So this is the required answer. Hence option (D) is the correct answer.

Answer 2

Explanation:

Let L be the length, r be the radius of the wire.

Volume of the wire is

V=πr

2

L

Differentiating both sides, we get

ΔV=π(2rΔr)L+πr

2

ΔL

As the volume of the wire remains unchanged when it gets stretched, so ΔV=0. Hence

0=2πrLΔr+πr

2

ΔL

∴

ΔL/L

Δr/r

=−

2

1

Poisson

′

sratio=

Longitudinalstrain

Lateralstrain

=−

ΔL/L

Δr/r

=

2

1

=0.5