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(B) In an Atwood machine,the tension $T$ in the string is given by $T = \frac{2 m_1 m_2 g}{m_1 + m_2}$. The extension $x$ in the spring is proportiona

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$x_2 > x_1 > x_3$

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(B) In an Atwood machine,the tension $T$ in the string is given by $T = \frac{2 m_1 m_2 g}{m_1 + m_2}$. The extension $x$ in the spring is proportional to the tension $T$ $(x = T/k)$.Case $1$: $m_1 = 2 \ kg, m_2 = 2 \ kg$. Tension $T_1 = \frac{2 	imes 2 	imes 2 	imes g}{2 + 2} = 2g$.Case $2$: $m_1 = 3 \ kg, m_2 = 2 \ kg$. Tension $T_2 = \frac{2 	imes 3 	imes 2 	imes g}{3 + 2} = 2.4g$.Case $3$: $m_1 = 1 \ kg, m_2 = 2 \ kg$. Tension $T_3 = \frac{2 	imes 1 	imes 2 	imes g}{1 + 2} = 1.33g$.Comparing the tensions,we get $T_2 > T_1 > T_3$.Since $x \propto T$,the extensions follow the same order: $x_2 > x_1 > x_3$.

The same spring is attached to $2 \ kg, 3 \ kg$ and $1 \ kg$ blocks in three different cases as shown in the figure. If $x_1, x_2$ and $x_3$ are the extensions in the spring in these cases,respectively,find the correct relation between them.

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