First slide

Elastic Modulie

Question

A copper wire of cross sectional area A is under a tension F. Young's modulus of copper is $γ$ and Poisson's ratio $σ .$ Then find the fractional change in its cross sectional area.

Moderate

A

$\frac{σ F}{4 A γ}$
B

$\frac{σ F}{2 A γ}$
C

$\frac{2 σ F}{A γ}$
D

$\frac{σ F}{A γ}$

Solution

(3)
$\begin{array}{l} S t r e s s S = F / A \\ L o n g i t u d i n a l s t r a i n ε = \frac{S}{γ} = \frac{F}{A γ} \\ N o w, \frac{Δ r / r}{Δ l / l} = - σ \Rightarrow \frac{Δ r}{r} = - σ . \frac{Δ l}{l} = - σ ε \\ N o w, A = π r^{2} \\ ∴ \frac{Δ A}{A} = 2. (\frac{Δ r}{r}) = - 2 σ ε \\ ∴ F r a c t i o n a l d e c r e a s e i n i t s c r o s s \sec t i o n a l a r e a \\ = 2 σ ε = \frac{2 σ F}{A Y} . \end{array}$

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