A small spherical ball is suspended through a string of length `l` . The whole arrangement is placed in a vehicle which is moving with velocity `v` . Now, suddenly the vehicle stops and ball starts moving along a circular path. If tension in the string at the highest point is twice the weight of the ball then (assume that the ball completes the vertical circle)

Options

A. `v=sqrt(5gl)`
B. `v=sqrt(7gl)`
C. velocity of the ball at highest point is `sqrt(gl)`
D. velocity of the ball at highest point is `sqrt(3gl)`

Explanation

To solve the problem step by step, we will analyze the motion of the ball and apply the principles of conservation of energy and centripetal force.

### Step 1: Understand the Setup
A small spherical ball is suspended from a string of length `l`. When the vehicle stops suddenly, the ball starts moving in a vertical circular path.

### Step 2: Define Variables
- Let the mass of the ball be `m`.
- The initial velocity of the ball when the vehicle stops is `v`.
- The velocity of the ball at the highest point of the circular path is `v'`.
- The tension in the string at the highest point is given as `T = 2mg`.

### Step 3: Apply Conservation of Energy
Using the conservation of energy between the point where the ball starts moving (point A) and the highest point of the circular path (point B):

\[ \frac{1}{2} mv^2 = \frac{1}{2} mv'^2 + mg(2l) \]

This equation states that the initial kinetic energy of the ball is equal to the sum of its kinetic energy at the highest point and the potential energy gained when it rises to a height of `2l`.

### Step 4: Simplify the Energy Equation
Rearranging the equation gives:

\[ \frac{1}{2} mv^2 = \frac{1}{2} mv'^2 + 2mgl \]

Dividing through by `m`:

\[ \frac{1}{2} v^2 = \frac{1}{2} v'^2 + 2gl \]

Multiplying through by 2:

\[ v^2 = v'^2 + 4gl \quad \text{(Equation 1)} \]

### Step 5: Analyze Forces at the Highest Point
At the highest point (point B), the forces acting on the ball are the tension in the string and the weight of the ball. The net force provides the centripetal force required for circular motion:

\[ T + mg = \frac{mv'^2}{l} \]

Substituting the given tension:

\[ 2mg + mg = \frac{mv'^2}{l} \]

This simplifies to:

\[ 3mg = \frac{mv'^2}{l} \]

Dividing through by `m`:

\[ 3g = \frac{v'^2}{l} \]

Rearranging gives:

\[ v'^2 = 3gl \quad \text{(Equation 2)} \]

### Step 6: Substitute Equation 2 into Equation 1
Now, substitute Equation 2 into Equation 1:

\[ v^2 = 3gl + 4gl \]

This simplifies to:

\[ v^2 = 7gl \]

Taking the square root gives:

\[ v = \sqrt{7gl} \]

### Step 7: Find the Velocity at the Highest Point
From Equation 2, we already found:

\[ v' = \sqrt{3gl} \]

### Conclusion
The initial velocity of the ball when the vehicle stops is \( v = \sqrt{7gl} \) and the velocity at the highest point is \( v' = \sqrt{3gl} \).

### Final Answers
- The initial velocity \( v = \sqrt{7gl} \)
- The velocity at the highest point \( v' = \sqrt{3gl} \)