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

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

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

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

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

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Solution

The correct option is A
$\frac{2 m g}{3 a Y}$

The acceleration of the system is

$a = \frac{F_{n e t}}{m_{t o t a l}} = \frac{m g}{3 m} = \frac{g}{3}$ .

Let $T_{1}$ and $T_{2}$ be the tensions in $w_{1}$ and $w_{2}$ . Then

$T_{1} = m a = \frac{m g}{3}$ .

Also,

$T_{2} - T_{1} = m a$

$\Rightarrow T_{2} = \frac{2 m g}{3}$

The strain in the wire $w_{2}$ is

$s t r a i n = \frac{T_{2}}{a Y} = \frac{2 m g}{3 a Y}$

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