CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
1347293_117b365127964373a1c4b5bbf886c6b0.png


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

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
$$ \begin{array}{l} \text { we define a quantity } \\ \qquad K=\frac{F}{\Delta L}=\frac{y_{A}}{L} \quad \therefore\left(y=\frac{F L}{A \Delta L}\right) \end{array} $$
$$ \text { in case of parral } $$
$$ \begin{array}{l} \text { Keq }=K_{1}+K_{2} \\ \frac{Y_{e q}(2 A)}{L_{1}}=\frac{Y_{1} A}{L}+\frac{Y_{2} A}{L} \\ \text { 2Yeq }=Y_{1}+Y_{2} \end{array} $$
$$ Y_{e q}=\frac{Y_{1}+Y_{2}}{2} $$
$$ \begin{array}{l} \text { Potential Energy }(U)=\frac{1}{2} \times \text { stress } x \text { strain } x \text { volume } \\ \qquad U=\frac{1}{2} \times \text { stress } \times \frac{\text { Stress }}{Y_{e q}} \times(2 A \times L) \quad \therefore Y=\frac{\text { Stress }}{\text { strain }} \end{array} $$
$$ \begin{array}{l} =\frac{1}{2}\left(\frac{F}{2 A}\right)^{2} \times \frac{1}{Y e q} \times(2 A \cdot L) \\ U=\frac{M^{2} g^{2} L}{2 A\left(Y_{1}+Y_{2}\right)} \end{array} $$
$$ \begin{array}{l} \therefore F=m g \\ \therefore Y_{e q}=\frac{Y_{1}+Y_{2}}{2} \\ \therefore v=2 A L \end{array} $$

1990444_1347293_ans_c67e7b0ee583496a92d670faf3aec121.png

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image