Question

In determination of young's modulus of elasticity of wire, a force is applied and extension is recorded. Initial length of wire is '$1\,m$'. The curve between extension and stress is depicted then Young's modulus of wire will be :-

• A. $2 \times 10^{9} N / m ^{2}$
• B. $1 \times 10^{9} N / m ^{2}$
• C. $2 \times 10^{10} N / m ^{2}$
• D. $1 \times 10^{10} N / m ^{2}$

Answer 1

Answer: (B)

Slope $=\frac{\Delta L }{ F / A }$
$=\frac{7 \times 10^{-3} m }{7000 \times 10^{3} N / m ^{2}}$
$\Delta L =\frac{ FL }{\dot{ A } Y } $
$\Rightarrow \frac{1}{ Y }=\frac{\Delta L }{\overline{ F } / A L }$
$=\frac{7 \times 10^{-3}}{7000 \times 10^{-3} \times 1} m ^{2} / N$
On solving $Y =1 \times 10^{9} N / m ^{2}$