Question

# A Plank of mass $$M$$ is suspended horizontally by using two wires as shown in the figure. They havesame length (L) and same cross-sectional area (A). Their Young's moduli are $$Y _ { 1 }$$ and $$Y _ { 2 }$$ respectively.The elastic potential energy of the system will be

A
2M2g2LA(Y1+Y2)
B
M2g2LA(Y1+Y2)
C
M2g2L2A(Y1+Y2)
D
M2g2L(Y1+Y2)2AY1Y2

Solution

## The correct option is B $$\dfrac { M ^ { 2 } g ^ { 2 } L } { 2 A \left( Y _ { 1 } + Y _ { 2 } \right) }$$- $$A=$$ Areas of cross section $$L=$$ length of wire $$Y_{1} Y_{2}=$$ Young modulus of $$M=$$mass of plank $$\begin{array}{l} \text { we define a quantity } \\ \qquad K=\frac{F}{\Delta L}=\frac{y_{A}}{L} \quad \therefore\left(y=\frac{F L}{A \Delta L}\right) \end{array}$$ $$\text { in case of parral }$$ $$\begin{array}{l} \text { Keq }=K_{1}+K_{2} \\ \frac{Y_{e q}(2 A)}{L_{1}}=\frac{Y_{1} A}{L}+\frac{Y_{2} A}{L} \\ \text { 2Yeq }=Y_{1}+Y_{2} \end{array}$$ $$Y_{e q}=\frac{Y_{1}+Y_{2}}{2}$$ $$\begin{array}{l} \text { Potential Energy }(U)=\frac{1}{2} \times \text { stress } x \text { strain } x \text { volume } \\ \qquad U=\frac{1}{2} \times \text { stress } \times \frac{\text { Stress }}{Y_{e q}} \times(2 A \times L) \quad \therefore Y=\frac{\text { Stress }}{\text { strain }} \end{array}$$ $$\begin{array}{l} =\frac{1}{2}\left(\frac{F}{2 A}\right)^{2} \times \frac{1}{Y e q} \times(2 A \cdot L) \\ U=\frac{M^{2} g^{2} L}{2 A\left(Y_{1}+Y_{2}\right)} \end{array}$$ $$\begin{array}{l} \therefore F=m g \\ \therefore Y_{e q}=\frac{Y_{1}+Y_{2}}{2} \\ \therefore v=2 A L \end{array}$$Physics

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