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Two wires of same materials (Young modulus) and same length L but radii R and 2R respoctirly are joined end to end and a weight W is suspended from the combination as shown in figure.

The elastic PE stored in the system is-

Option: 1

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


Option: 2

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


Option: 3

\frac{3 w^2 L}{8 \pi_R^2 y}


Option: 4

\frac{\omega^2 L}{8 \pi R^2 y}


Answers (1)

best_answer

Using, 

\quad U=\frac{1}{2} k x^2=\frac{k}{2}\left(\frac{F}{k}\right)^2=\frac{F^2}{2k}
We have 

\begin{aligned} U&=U_1+U_2\\ & =\frac{F^2}{2 k_1}+\frac{F^2}{2 k_2} \quad(F=W) \\ & =\frac{W^2}{2}\left[\frac{1}{k_1}+\frac{1}{k_2}\right]~~~~~~~~...(i) \\ \text{and}\\ k_1&=\frac{Y A}{L} =\frac{Y \pi(2 R)^2}{L}, \\k_2&=\frac{Y \pi R^2}{L} \end{aligned}

Put the raluers k_1 ~\& ~k_2 in equation (i)

\begin{aligned} & U=\frac{W^2}{2}\left[\frac{L}{4 \pi R^2 Y}+\frac{L}{\pi R^2 Y}\right] \\ & U=\frac{5 \omega^2 L}{8 \pi R^2 Y} \end{aligned}

Posted by

Deependra Verma

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