Question

Two wires of the same material (young’s modulus Y) and same length L but radii R and 2R respectively are joined end to end and a weight W is suspended from the combination as shown in the figure. The elastic potential energy in the system in equilibrium is:

A) ${\displaystyle \frac{3W^{2}L}{4\pi R^{2}Y}}$.
B) ${\displaystyle \frac{3W^{2}L}{8\pi R^{2}Y}}$.
C) ${\displaystyle \frac{5W^{2}L}{8\pi R^{2}Y}}$.
D) ${\displaystyle \frac{W^{2}L}{\pi R^{2}Y}}$.

Answer 1

Hint: The potential energy is the energy which is saved inside the wire and is caused due to the elongation of the wire. The young’s modulus of a material is defined as the ratio of the stress and strain and it is constant till the proportional limit.

Formula used: The formula of the potential energy is given by,
$\Rightarrow E={\displaystyle \frac{F^{2}L}{2AY}}$
Where force is F, the original length is L, the area is A and Young’s modulus is Y.

Complete step by step solution:
It is given in the problem that the two wires of the same material (young’s modulus Y) and same length L but radii R and 2R respectively are joined end to end and a weight W is suspended from the combination and we need to find the elastic potential energy in the system in equilibrium condition.
The formula of the potential energy is given by,
$\Rightarrow E={\displaystyle \frac{F^{2}L}{2AY}}$
Where force is F, the original length is L, the area is A and Young’s modulus is Y.
The potential energy of the wire is equal to,
$\Rightarrow E={\displaystyle \frac{F^{2}L}{2AY}}$
The elongation is taking place in the two parts of the wires.
The potential energy of the system is equal to,
$\Rightarrow E={\displaystyle \frac{W^{2}L}{2A_{1}Y}}+{\displaystyle \frac{W^{2}L}{2A_{2}Y}}$
$\Rightarrow E={\displaystyle \frac{W^{2}L}{2\left(\pi R^2\right)Y}}+{\displaystyle \frac{W^{2}L}{2\pi {\left(2R\right)}^{2}Y}}$
$\Rightarrow E={\displaystyle \frac{W^{2}L}{2\pi R^{2}Y}}+{\displaystyle \frac{W^{2}L}{8\pi R^{2}Y}}$
$\Rightarrow E={\displaystyle \frac{2W^{2}L+W^{2}L}{8\pi R^{2}Y}}$
$\Rightarrow E={\displaystyle \frac{3W^{2}L}{8\pi R^{2}Y}}$.

The potential energy of the wire is equal to $E={\displaystyle \frac{3W^{2}L}{8\pi R^{2}Y}}$. The correct option for this problem is option B.

Note: The students are advised to understand and remember the formula of the potential energy of the wire as it is very useful in solving these kinds of problems. The change of the length happens due to the applied force or the load due to weight.