Question

A load of $2 \,kg$ produces an extension of $1 \,mm$ in a wire of $3\, m$ in length and $1\, mm$ in diameter. The Young's modulus of wire will be

• A. $3.25 \times 10^{10} Nm ^{-2}$
• B. $7.48 \times 10^{12} Nm ^{2}$
• C. $7.48 \times 10^{10} Nm ^{-2}$
• D. $7.48 \times 10^{-10} Nm ^{-2}$

Answer 1

Answer: (C)

We know
$\frac{\text { Force } \times \text { Length }}{\text { Area of cross-section } \times \text { elongation }}=\text { Young's Modulus } $
$=\frac{F \times L}{A \times \Delta L}=Y $
$\{F=2 \times 10\, N , A=\pi \times(1 / 2)^{2} \times 10^{-6} m ^{2} ,
L=3\, m \, \Delta L=1 \times 10^{-3} m\}$
Substituting values
$\frac{20 \times 3}{\pi \times \frac{1}{4} \times 10^{-6} \times 1 \times 10^{-3}}=Y $
$\frac{20 \times 3 \times 4}{3.14 \times 10^{-9}}=Y $
$7.48 \times 10^{10} Nm ^{-2}=Y$