Three masses \(m_1 = 4\) kg, \(m_2 = 4\) kg and \(m_3 = 6\) kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of \(T_1/T_2\) is _____.
(take \(g = 10\) m/s\(^2\))

Let the acceleration of the system be \(a\). The total mass on the right side is \(m_2 + m_3 = 4 + 6 = 10\) kg, and the mass on the left side is \(m_1 = 4\) kg. Since \(10\) kg \(> 4\) kg, the right side accelerates downwards and the left side accelerates upwards.

The common acceleration \(a\) of the system is given by:
\(a = \dfrac{(m_2 + m_3) - m_1}{m_1 + m_2 + m_3} g\)
\(a = \dfrac{10 - 4}{4 + 4 + 6} g = \dfrac{6}{14} g = \dfrac{3}{7} g\)

For mass \(m_1\), the equation of motion is:
\(T_1 - m_1 g = m_1 a\)
\(T_1 = m_1(g + a) = 4 \left(g + \dfrac{3}{7}g\right) = 4 \left(\dfrac{10}{7}g\right) = \dfrac{40}{7}g\)

For mass \(m_3\), the equation of motion is:
\(m_3 g - T_2 = m_3 a\)
\(T_2 = m_3(g - a) = 6 \left(g - \dfrac{3}{7}g\right) = 6 \left(\dfrac{4}{7}g\right) = \dfrac{24}{7}g\)

The ratio of the tensions is:
\(\dfrac{T_1}{T_2} = \dfrac{\dfrac{40}{7}g}{\dfrac{24}{7}g} = \dfrac{40}{24} = \dfrac{5}{3}\)

Answer: \(5/3\)