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Q.

Two wires of the same material (Young’s modulus = Y) and same length L but radii R and 3R 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

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a

$\frac{3 w^{2} L}{4 π R^{2} Y}$

b

$\frac{10 w^{2} L}{18 π R^{2} Y}$

c

$\frac{5 w^{2} L}{8 π R^{2} Y}$

d

$\frac{w^{2} L}{π R^{2} Y}$

answer is B.

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Detailed Solution

$\begin{array}{l} Δ l_{1} = \frac{w L}{9 π R^{2} Y}, Δ l_{2} = \frac{w L}{π R^{2} Y} \\ ∴ U = \frac{1}{2} K_{1} Δ {l_{1}}^{2} + \frac{1}{2} K_{2} Δ {l_{2}}^{2} (h e r e K = \frac{Y A}{L}) \\ = \frac{1}{2} \times \frac{Y 9 π R^{2}}{L} \times {(\frac{w L}{9 π R^{2} Y})}^{2} + \frac{1}{2} \times \frac{Y π R^{2}}{L} \times {(\frac{w L}{π R^{2} Y})}^{2} \\ = \frac{10 w^{2} L}{18 π R^{2} Y} \end{array}$

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