One end of massless inextensible string of length `l` is fixed and other end is tied to a small ball of mass `m`. The ball is performing a circular motion in vertical plane. At the lowest position, speed of ball is `sqrt(20gl)`. Neglect any other forces on the ball except tension and gravitational force. Acceleration due to gravity is `g`.
At the highest position of ball, tangential acceleration of ball is `-`

Options

A. 0
B. `g`
C. `5g`
D. `16g`

Explanation

To find the tangential acceleration of the ball at the highest position in its circular motion, we can follow these steps:

### Step 1: Understand the problem
We have a ball of mass `m` attached to a massless, inextensible string of length `l`. The ball is performing circular motion in a vertical plane. At the lowest position, the speed of the ball is given as \( v = \sqrt{20gl} \).

### Step 2: Use the conservation of energy
At the lowest point, the potential energy (PE) is zero, and the kinetic energy (KE) is given by:
\[ KE_{low} = \frac{1}{2} m v^2 = \frac{1}{2} m (20gl) = 10 mgl \]

At the highest point, the height of the ball is \( 2l \), so the potential energy at the highest point is:
\[ PE_{high} = mgh = mg(2l) = 2mgl \]

Let \( v_{high} \) be the speed of the ball at the highest point. The kinetic energy at the highest point is:
\[ KE_{high} = \frac{1}{2} m v_{high}^2 \]

By the conservation of mechanical energy:
\[ KE_{low} = KE_{high} + PE_{high} \]
Substituting the values:
\[ 10 mgl = \frac{1}{2} m v_{high}^2 + 2mgl \]

### Step 3: Solve for \( v_{high} \)
Rearranging the equation:
\[ 10 mgl - 2mgl = \frac{1}{2} m v_{high}^2 \]
\[ 8 mgl = \frac{1}{2} m v_{high}^2 \]
Dividing both sides by \( m \):
\[ 8gl = \frac{1}{2} v_{high}^2 \]
Multiplying by 2:
\[ 16gl = v_{high}^2 \]
Taking the square root:
\[ v_{high} = \sqrt{16gl} = 4\sqrt{gl} \]

### Step 4: Find the centripetal acceleration
At the highest point, the forces acting on the ball are the gravitational force \( mg \) acting downward and the tension \( T \) in the string also acting downward. The net force towards the center (which provides centripetal acceleration) is:
\[ F_{net} = mg + T \]
This net force must equal the centripetal force:
\[ F_{net} = \frac{mv_{high}^2}{l} \]
Substituting \( v_{high}^2 \):
\[ mg + T = \frac{m(16gl)}{l} \]
Simplifying:
\[ mg + T = 16mg \]
Thus:
\[ T = 16mg - mg = 15mg \]

### Step 5: Determine the tangential acceleration
Tangential acceleration is related to the change in speed along the path of motion. At the highest point, the only forces acting are the gravitational force and tension, and since the ball is moving in a circular path, there is no net force acting tangentially to the motion. Therefore, the tangential acceleration is:
\[ a_{tangential} = 0 \]

### Final Answer
The tangential acceleration of the ball at the highest position is \( 0 \).

---