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 is :

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

Answer 1

The correct option is C $\frac{5w^{2}L}{8\pi R^{2}Y}$

Tension in both the wires is same. Elongation in the first wire is $e_1=\frac{\text{WL}}{\pi R^{2}Y}Energyinthefirstwireis{U}_1=\frac{1}{2}w\times e_1=\frac{w^{2}L}{2\pi R^{2}Y}Elongationinthesecondwireis{e}_2=\frac{WL}{\pi {\left(2R\right)}^{2}Y}Energyinthesecondwireis{U}_2=\frac{1}{2}w\times e_2=\frac{w^{2}L}{8\pi R^{2}Y}\therefore Totalenergy\text{Total energy}=U_1+U_2=\frac{5W^{2}L}{8\pi R^{2}Y}$