Discussion Forum

 The Young's  modulus of a material $2 \times 10^{11}$ N/m² 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 

Answer:

Explanation:

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 ,

    $Y= \frac{stress}{Strain}\Rightarrow Y= \frac{stress}{\frac{\triangle L}{l}}$

Here, l= original length of the wire,

ΔL= charge in the length of the wire

ΔL = $\frac{Stress \times  l}{Y}$=$\frac{1\times 10^{8}1}{2\times 10^{11}}$

 = $0.5 \times 10^{-3}$=0.5 mm