Problem Statement:
A Bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a horizontal velocity V1 at the lowest point A such that it completes a semicircular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point C. Obtain an expression for V1 and the speed at C.

Solution:
To solve this problem, we need to analyze the forces acting on the bob at different points on its trajectory.

1. Forces at Point A:
- At the lowest point A, the bob is at its maximum height from the vertical plane. The only force acting on the bob is the tension in the string, which is directed towards the center of the circular path.
- The weight of the bob acts vertically downwards, but it does not cause any change in the motion of the bob at this point.

2. Forces at Point B:
- As the bob moves along the circular path, it reaches a point B where the string is at an angle with the vertical.
- At this point, the tension in the string can be resolved into two components: one along the circular path (Tcosθ) and the other perpendicular to the circular path (Tsinθ).
- The weight of the bob acts vertically downwards, but it does not cause any change in the motion of the bob at this point.

3. Forces at Point C:
- At the topmost point C, the string becomes slack and the tension in the string becomes zero.
- The only force acting on the bob at this point is its weight, which acts vertically downwards.
- The weight of the bob causes it to accelerate downwards, and the bob loses its circular motion at this point.

Expression for V1:
- At point A, the tension in the string is equal to the centripetal force required to keep the bob in circular motion.
- The centripetal force is given by the equation Fc = mv²/r, where m is the mass of the bob, v is the velocity, and r is the radius of the circular path.
- At point A, the radius of the circular path is equal to the length of the string, so r = L.
- Therefore, the tension in the string at point A is T = mv²/L.
- Since the bob is imparted a horizontal velocity V1 at point A, we can equate the tension in the string to the centripetal force and solve for V1.
- T = mv²/L = mV1²/L
- V1² = gL
- V1 = √(gL)

Speed at Point C:
- At point C, the tension in the string becomes zero and the bob loses its circular motion.
- The only force acting on the bob at this point is its weight, which causes it to accelerate downwards.
- The speed of the bob at point C can be determined using the equation for free fall motion: v = √(2gh), where g is the acceleration due to gravity and h is the height from which the bob falls.
- At point C, the height from which the bob falls is equal to the length of the string, so h = L.
- Therefore, the speed at point C is v = √(2gL).

Conclusion: