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Two wires of same material 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-

Option: 1

\frac{3 w^2 L}{4 \pi R^2 Y}


Option: 2

\frac{3 w^2 L}{8 \pi R^2 Y}


Option: 3

\frac{5 w^2 L}{8 \pi R^2 Y}


Option: 4

\frac{w^2 L}{\pi R^2 Y}


Answers (1)

best_answer

\begin{aligned} & \Delta l_1=\frac{w L}{\left(4 \pi R^2\right) Y}, \, \Delta l_2=\frac{w L}{\pi R^2 Y} \\ & \therefore \quad U=\frac{1}{2} k_1\left(\Delta l_1\right)^2+\frac{1}{2} k_2\left(\Delta l_2\right)^2 \\ & U=\frac{1}{2} \times Y\left(\frac{4 \pi R^2}{L}\right) \times\left[\frac{w L}{4 \pi R^2 Y}\right]^2+\frac{1}{2} \times \frac{Y\left(\pi R^2\right)}{L} \times\left[\frac{w L}{\pi R^2 Y}\right]^2 \\ & U=\frac{5 w^2 L}{8 \pi R^2 Y} \\ & \end{aligned}

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