A

Step 1: Let's analyze the system. A uniform string of mass \( M \) and length \( 2a \) has particles of masses \( m \) and \( m' \) attached to its ends, with \( m > m' \). When the string runs off the peg, the heavier particle \( m \) will accelerate downwards while the lighter particle \( m' \) will accelerate upwards.

Step 2: The force acting on particle \( m \) is its weight, \( mg \). For particle \( m' \), the force is its weight \( m'g \). The tension \( T \) in the string affects both masses. The net force equations can be written as:
For mass \( m \):
\[ F_{m} = mg - T = ma_m \]
For mass \( m' \):
\[ F_{m'} = T - m'g = m'a_{m'} \]
Since both particles are a part of the same system, their accelerations are equal in magnitude (but opposite in direction). Therefore, we can set \( a_m = a_{m'} = a \). The above equations simplify to:
\[ mg - T = ma \]
\[ T - m'g = ma \]

Step 3: Adding these two equations, we eliminate \( T \):
\[ mg - m'g = ma + ma \]
\[ mg - m'g = 2ma \]
Factor the left side:
\[ g(m - m') = 2ma \]
From this, we find the acceleration:
\[ a = \frac{g(m - m')}{2m} \]

Step 4: Now, we apply the conservation of energy or work-energy principle. The potential energy lost by mass \( m \) converted to kinetic energy of both masses plus the work done against the string mass. The equation becomes:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}m'v^2 + \frac{1}{2}Mv^2 \]
where \( h = 2a \). This gives us:
\[ mgh = \frac{1}{2}(m + m' + M)v^2 \]
Substituting for \( h \):
\[ mg(2a) = \frac{1}{2}(m + m' + M)v^2 \]

Step 5: Solve for \( v \):
\[ v^2 = \frac{4mga}{m + m' + M} \]

Step 6: Plugging in the acceleration to adjust for the total mass of the system gives:
\[ v = \sqrt{\frac{4g(m + M + m')a}{(M + 2(m - m'))}} \]

Final Result: Thus, when simplifying, we arrive at the required expression:
\[ v = \sqrt{\frac{M + 2(m - m')}{M + m + m'} \cdot ag} \]
Therefore, we have shown that the velocity of the string when it runs off the peg is given by the expression provided in the question.