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

\frac{3 m g}{4 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 $\frac{3 m g}{4 Y}$

If the system moves with acceleration a and T is the tension in the string $W_{2}$ then by comparing this condition from standard case $T = \frac{m_{1} m_{2}}{m_{1} + m_{2}} g$
In the given problem $m_{1} = (m + m) = 3 m$ and $m_{2} = m$
$∴ T e n s i o n = \frac{m 3 m g}{m + 3 m} = \frac{3}{4} m g$
$∴ S t r e s s = \frac{T}{a} = \frac{3}{4 a} m g$ and $S t r a i n = \frac{S t r e s s}{Y o u n g^{'} s m o d u l u s} = \frac{3 m g}{4 a Y}$

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Three blocks, each of same mass m, are connected with wires W1 and W2 of same cross-sectional area a and Young’s modulus Y. Neglecting friction between blocks and the table, the strain developed in wire W2 is

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