Step-by-Step Solution:

1. Analyze the Geometry of the System Let's consider the system at an instant when the separation between the two particles is 2x. The midpoint of the string, M, has been pulled up by a vertical distance y.

   • The total length of the string is 2a.
   • The length of the string from the midpoint M to each particle is a.
   • The horizontal distance of each particle from the center line is x.

   A right-angled triangle is formed with the string segment of length a as the hypotenuse, the horizontal distance x as the base, and the vertical displacement y as the height.

   From Pythagoras' theorem, we have the constraint equation: $x^2 + y^2 = a^2$ $y = \sqrt{a^2 - x^2}$

   Let θ be the angle the string makes with the horizontal surface. From the triangle, we can write: $\sin(\theta) = {y \over a}$ $\cos(\theta) = {x \over a}$

2. Analyze the Forces at the Midpoint The midpoint of the string M is being pulled upwards by a constant vertical force F. There are two tension forces, T, acting downwards along each segment of the string.

   Since the string is light (massless), the net force on the midpoint M must be zero (otherwise, it would have infinite acceleration). We consider the balance of forces in the vertical direction at point M: $F = T\sin(\theta) + T\sin(\theta) = 2T\sin(\theta)$

   From this, we can express the tension T in the string: $T = {F \over {2\sin(\theta)}}$

3. Analyze the Forces on a Particle Consider one of the particles of mass m. It moves on a frictionless horizontal surface. The forces acting on it are:

   • Weight mg (downwards)
   • Normal force N from the surface (upwards)
   • Tension T along the string.

   The vertical forces mg and N cancel each other out. The motion is purely horizontal. The net horizontal force is the horizontal component of the tension, $T\cos(\theta)$, which pulls the particle towards the center.

   According to Newton's second law of motion, this net horizontal force equals mass times acceleration (acc): $T\cos(\theta) = m \cdot acc$

4. Derive the Acceleration Now we can substitute the expression for T from Step 2 into the equation from Step 3: $\left( {F \over {2\sin(\theta)}} \right) \cos(\theta) = m \cdot acc$ ${F \over 2} {\cos(\theta) \over \sin(\theta)} = m \cdot acc$ ${F \over 2} \cot(\theta) = m \cdot acc$

   Solving for the acceleration acc: $acc = {F \over {2m}} \cot(\theta)$

5. Express Acceleration in Terms of x and a From the geometry in Step 1, we can find an expression for $\cot(\theta)$: $\cot(\theta) = {\text{adjacent} \over \text{opposite}} = {x \over y}$

   We also know that $y = \sqrt{a^2 - x^2}$. Substituting this into the expression for $\cot(\theta)$: $\cot(\theta) = {x \over {\sqrt{a^2 - x^2}}}$

   Finally, substitute this expression for $\cot(\theta)$ back into our equation for acceleration: $acc = {F \over {2m}} \left( {x \over {\sqrt{a^2 - x^2}}} \right)$

   So, the magnitude of the acceleration is: $acc = {F \over {2m}} {x \over {\sqrt{a^2 - x^2}}}$

6. Compare with Options The derived expression for acceleration matches option B.

   • A: ${F \over {2m}}{a \over {\sqrt {{a^2} - {x^2}} }}$ - Incorrect.
   • B: ${F \over {2m}}{x \over {\sqrt {{a^2} - {x^2}} }}$ - Correct.
   • C: ${F \over {2m}}{x \over a}$ - Incorrect.
   • D: ${F \over {2m}}{{\sqrt {{a^2} - {x^2}} } \over x}$ - Incorrect.