Question

The Young’s modulus of the material of a wire is 2 x 10¹⁰ N/m². If the elongation strain is 1%, then the energy stored in the wire per unit volume in J/m³ is :
1. 10⁶
2. 10⁸
3. 2 × 10⁸
4. 0.5 × 10⁶

Answer 1

Correct Answer - Option 1 : 10⁶

Concept:

• Elastic potential energy: When we exert tensile stress on a wire, it will get stretched and work done in stretching the wire will be equal and opposite to the work done by inter-atomic restoring force. This work stored in the wire in a form of Elastic potential energy.
• Now by using the relation of Young’s Modulus we can say that, 

 

​\(Y = \frac{F}{A} × \frac{L}{{\rm{l}}} \Rightarrow F = \frac{{YAl}}{L}\)

Substituting this in the equation of work we get

\(W = \smallint \frac{{YAl}}{L}dl = \frac{{YA{l^2}}}{{2L}} = \frac{1}{2} × Y × {\left( {\frac{l}{L}} \right)^2} × LA\)

\(\therefore W = Young's\;modulus × strai{n^2} × Volume\;of\;wire\;\)

• Hence work done per unit volume is given as

\(U = \frac{W}{V} = \frac{1}{2} × Young's\;modulus × strai{n^2}\)

\(also,\;U = \frac{1}{2} × stress × strain = \frac{1}{2} × \sigma\epsilon \)

Whereas Young’s modulus, \(Y = \frac{{stress}}{{strain}}\)

Using this the above equation can be modified as

\(U = \frac{{strain\;energy}}{{volume}} = \frac{1}{2} × Y × strai{n^2}\).......(i)

Explanation:

Given that, Y = 2 x 1010 N/m²,elongation strain is 1%,U  = energy stored per unit volume.

Now as we have all the values just put it in the equation number (i)

U = (1/2) × 2 x 10¹⁰ × (1/100)² = 10⁶ J/m³.

So energy stored in the wire per unit volume in J/m3 is 10⁶.