Three blocks, each of same mass $m$ , are connected with wire $W_{1}$ and $W_{2}$ of same cross sectional area 'a' and Young's modulus Y. Neglecting friction, the strain developed in wire $W_{2}$ is

\frac{2}{3} \frac{m g}{a Y}

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\frac{3}{2} \frac{m g}{a Y}

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\frac{1}{3} \frac{m g}{a Y}

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\frac{3 m g}{a Y}

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Solution

The correct option is A $\frac{2}{3} \frac{m g}{a Y}$
Writing equations of motion for all three particles
$m g - T = m a_{1}$
$T - T_{1} = m a_{2}$
$T_{1} = m a_{3}$
$∴ T = 2 m a$
$m g - 2 m a = m a$
$m g = 3 m a$
$a = g / 3$
$∴ T = 2 m \frac{g}{3}$
$Y = \frac{F / A}{△ L / L}$
$\frac{△ L}{L} = s t r a i n = \frac{2 m g / 3}{Y} = \frac{2 m g}{3 a Y}$

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