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(B) For an Atwood machine with masses $m_1$ and $m_2$ (where $m_2 > m_1$),the acceleration $a$ is given by the formula:$a = \left(\frac{m_2 - m_1}{m_1

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$9 : 7$

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(B) For an Atwood machine with masses $m_1$ and $m_2$ (where $m_2 > m_1$),the acceleration $a$ is given by the formula:$a = \left(\frac{m_2 - m_1}{m_1 + m_2}\right) g$Given that the acceleration $a = \frac{g}{8}$,we substitute this into the equation:$\frac{g}{8} = \left(\frac{m_2 - m_1}{m_1 + m_2}\right) g$$\frac{1}{8} = \frac{m_2 - m_1}{m_1 + m_2}$Cross-multiplying gives:$m_1 + m_2 = 8(m_2 - m_1)$$m_1 + m_2 = 8m_2 - 8m_1$Rearranging the terms to group $m_1$ and $m_2$:$m_1 + 8m_1 = 8m_2 - m_2$$9m_1 = 7m_2$Therefore,the ratio of the masses is:$\frac{m_2}{m_1} = \frac{9}{7}$

$A$ light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (vertically). If the acceleration of the system is $(\frac{g}{8})$,then the ratio of masses $m_2 : m_1$ is:

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