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Same spring is attached with $2 \mathrm{~kg}, 3 \mathrm{~kg}$ and $1 \mathrm{~kg}$ blocks in three different cases as shown in figure. If $x_1, x_2, x_3$ be the extensions in the spring in the three cases, then
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The correct answer is:
$x_2>x_1>x_3$
If $T_1, T_2, T_3$ are the tensions in the strings in the three cases, we have
$\begin{aligned} & T_1=\frac{2 m_1 m_2 g}{m_1+m_2}=\frac{2 \times 2 \times 2 g}{(2+2)}=2 g \\ & T_2=\frac{2 \times 3 \times 2 g}{(3+2)}=2.4 g \\ & T_3=\frac{2 \times 1 \times 2 g}{(1+2)}=1.33 g \\ & \text { As } x \propto T \text { and } T_2>T_1>T_3 \\ & \therefore \quad x_2>x_1>x_3\end{aligned}$
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