The velocity at which \(6\) kg mass (shown in figure) strikes the ground when it is released from a height of \(6\) m above the ground is __________ m/s. Assume pulley is massless and string is light and inextensible. (Take g \(= 10\) m/s\(^2\))

Let the acceleration of the system be \(a\) and the tension in the string be \(T\).

The equations of motion for the two masses are:
\(6g - T = 6a\)
\(T - 2g = 2a\)

Adding the two equations, we get:
\(4g = 8a\)
\(\Rightarrow a = \dfrac{4g}{8} = \dfrac{g}{2}\)

Substituting \(g = 10 \text{ m/s}^2\), the acceleration is:
\(a = \dfrac{10}{2} = 5 \text{ m/s}^2\)

The \(6 \text{ kg}\) mass is released from rest (\(u = 0\)) and travels a downward distance \(s = 6 \text{ m}\).

Using the kinematic equation \(v^2 = u^2 + 2as\):
\(v^2 = 0 + 2 \times 5 \times 6\)
\(v^2 = 60\)
\(v = \sqrt{60} \approx 7.74 \text{ m/s}\)

Answer: \(7.74\)