Ratio of Tensions in Mass Pulley System

To find the ratio of the tensions $T_1/T_2$, we analyze the motion of the three-mass system connected over the fixed frictionless pulley.

1. Total Acceleration of the System

Let $m_1 = 4\text{ kg}$, $m_2 = 4\text{ kg}$, and $m_3 = 6\text{ kg}$.

• Total mass on the left side: $m_{\text{left}} = m_1 = 4\text{ kg}$
• Total mass on the right side: $m_{\text{right}} = m_2 + m_3 = 4\text{ kg} + 6\text{ kg} = 10\text{ kg}$

Since $m_{\text{right}} > m_{\text{left}}$, the system accelerates with the right side moving downward and the left side moving upward with an acceleration $a$:

$a = \left( \frac{m_{\text{right}} - m_{\text{left}}}{m_{\text{right}} + m_{\text{left}}} \right) g = \left( \frac{10 - 4}{10 + 4} \right) g = \frac{6}{14} g = \frac{3}{7} g$

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2. Tension $T_1$

Considering the vertical motion of mass $m_1$, which accelerates upwards:

$T_1 - m_1 g = m_1 a$

$T_1 = m_1 (g + a) = 4 \left( g + \frac{3}{7} g \right) = 4 \left( \frac{10}{7} g \right) = \frac{40}{7} g$

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3. Tension $T_2$

Considering the vertical motion of mass $m_3$, which accelerates downwards:

$m_3 g - T_2 = m_3 a$

$T_2 = m_3 (g - a) = 6 \left( g - \frac{3}{7} g \right) = 6 \left( \frac{4}{7} g \right) = \frac{24}{7} g$

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4. Ratio $T_1/T_2$

$\frac{T_1}{T_2} = \frac{\frac{40}{7} g}{\frac{24}{7} g} = \frac{40}{24} = \frac{5}{3}$

Thus, the value of $T_1/T_2$ is $\frac{5}{3}$, which corresponds to option A.