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Young's Modulus of Elasticity

Two blocks of...

Question

Two blocks of masses $m_{1}$ and $m_{2}$ are joined by a wire of Young’s modulus Y via a massless pulley. The area of cross-section of the wire is S and its length is L. When the system is released, increase in length of the wire is

\frac{m_{1} m_{2} g L}{Y S (m_{1} + m_{2})}

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\frac{2 m_{1} m_{2} g L}{Y S (m_{1} + m_{2})}

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\frac{(m_{1} - m_{2}) g L}{Y S (m_{1} + m_{2})}

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\frac{4 m_{1} - m_{2} g L}{Y S (m_{1} + m_{2})}

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Solution

The correct option is B $\frac{2 m_{1} m_{2} g L}{Y S (m_{1} + m_{2})}$
Tension in the wire $T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}}$ g $∴$ stress in the wire = $\frac{T}{S}$ = $\frac{2 m_{1} m_{2} g}{S (m_{1} + m_{2})}$
$∴$ Strain $\frac{l}{L}$ = $\frac{S t r e s s}{Y}$ = $\frac{2 m_{1} m_{2} g}{Y S (m_{1} + m_{2})}$ $\Rightarrow$ $l = \frac{2 m_{1} m_{2} g L}{Y S (m_{1} + m_{2})}$

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