Question 69: Properties of Matter

Question

An aluminium rod with Young's modulus $$Y=7.0 	imes 10^{10} \mathrm{~N} / \mathrm{m}^{2}$$ undergoes elastic strain of $$0.04 \%$$. The energy per unit volume stored in the rod in SI unit is:

Accepted Answer

The strain energy stored per unit volume in a material under stress can be calculated using the following formula:
$U = \frac{1}{2} \sigma \epsilon$
where $\sigma$ is the stress and $\epsilon$ is the strain.
For an elastic material, stress is proportional to strain (Hooke's law), and the constant of proportionality is the Young's modulus (Y). So we can write:
$\sigma = Y \epsilon$
Substituting this into the energy density equation we get:
$U = \frac{1}{2} Y \epsilon^2$
The strain given in the problem is 0.04%, which needs to be converted to a decimal for use in this formula. Therefore, $\epsilon = 0.04/100 = 0.0004$.
Substituting the values into the equation gives:
$U = \frac{1}{2} 	imes 7.0 	imes 10^{10} N/m^2 	imes (0.0004)^2 = 5600 ~J/m^3$

Answer

5600

Answer

2800

Answer

11200

Answer

8400

Properties of Matter

An aluminium rod with Young's modulus $Y=7.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} undergoes elastic strain of 0.04 \%$. The energy per unit volume stored in the rod in SI unit is:

Choose the correct answer: