A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g/8 then the ratio of the masses is

The correct option is 9 : 7.

Let us consider a system of two masses $m_1$ and $m_2$ connected by a light string passing over a smooth light pulley. Without loss of generality, let us assume that $m_1>m_2$.

When the system is released, the heavier mass $m_1$ accelerates downwards with acceleration $a$, and the lighter mass $m_2$ accelerates upwards with the same acceleration $a$.
The tension in the string is $T$.

Writing the equations of motion for both masses:
For mass $m_1$ moving downwards:
$m_{1}g-T=m_{1}a$ (Equation 1)

For mass $m_2$ moving upwards:
$T-m_{2}g=m_{2}a$ (Equation 2)

Adding Equation 1 and Equation 2 gives:
$(m_1-m_2)g=(m_1+m_2)a$

Therefore, the acceleration of the system is given by:
$a=\frac{m_1-m_2}{m_1+m_2}g$

Given that the acceleration of the system is $a=\frac{g}{8}$, we can substitute this value into the equation:
$\frac{g}{8}=\frac{m_1-m_2}{m_1+m_2}g$

Dividing both sides by $g$:
$\frac{1}{8}=\frac{m_1-m_2}{m_1+m_2}$

Cross-multiplying to solve for the ratio of the masses:
$m_1+m_2=8(m_1-m_2)$
$m_1+m_2=8m_1-8m_2$

Rearranging the terms:
$m_2+8m_2=8m_1-m_1$
$9m_2=7m_1$

Finding the ratio of the masses:
$\frac{m_1}{m_2}=\frac{9}{7}$

Thus, the ratio of the masses $m_1:m_2$ is 9 : 7.