Question

The Young’s modulus of a material is $2 \times 10^{11} N / m^2$ and its elastic limit is $1 \times 10^8 \, N /m^2$ . For a wire of $1\,m$ length of this material, the maximum elongation achievable is

• A. 0.2 mm
• B. 0.3 mm
• C. 0.4 mm
• D. 0.5 mm

Answer 1

Answer: (D)

Given,
Young's modulus of material $=2 \times 10^{11} N / m ^{2}$
Elastic limit or stress $=1 \times 10^{8} N / m ^{2}$
Length of the wire $=1 m$ We know that,
$Y=\frac{\text { Stress }}{\text { Strain }} \Rightarrow Y=\frac{\text { Stress }}{\frac{\Delta L}{l}}$
Here, $l=$ original length of the wire
$\Delta L=$ charge in the length of the wire
$\Delta L=\frac{\text { Stress } \times l}{Y}=\frac{1 \times 10^{8} \times 1}{2 \times 10^{11}}$
$=0.5 \times 10^{-3}=0.5 \,mm$