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`.
During circular motion, minimum value of tension in the string`-`

Options

A. zero only if `m_(1)=m_(2)`
B. `mg`
C. `10mg`
D. `15mg`

Explanation

To find the minimum value of tension in the string when the ball is performing circular motion, we can follow these steps:

### Step 1: Identify the forces acting on the ball at the highest position
At the highest point of the circular motion, the forces acting on the ball are:
- The gravitational force (weight) acting downward, which is \( mg \).
- The tension in the string, which also acts downward at this point.

### Step 2: Apply the centripetal force equation
At the highest point, the net force acting towards the center of the circular path must equal the centripetal force required to keep the ball moving in a circle. Therefore, we can write:
\[ T + mg = \frac{mv^2}{L} \]
where \( T \) is the tension in the string, \( m \) is the mass of the ball, \( v \) is the velocity at the highest point, and \( L \) is the length of the string.

### Step 3: Use conservation of energy to find the velocity at the highest point
Using the conservation of mechanical energy, we can relate the kinetic energy at the lowest point to the potential energy at the highest point:
- At the lowest point, the potential energy is 0, and the kinetic energy is \( \frac{1}{2} m u^2 \), where \( u = \sqrt{20gl} \).
- At the highest point, the potential energy is \( mg(2L) \) (since the height is \( 2L \)), and the kinetic energy is \( \frac{1}{2} m v^2 \).

Setting the total energy at the lowest point equal to the total energy at the highest point:
\[ \frac{1}{2} m u^2 = mg(2L) + \frac{1}{2} m v^2 \]
Substituting \( u^2 = 20gl \):
\[ \frac{1}{2} m (20gl) = mg(2L) + \frac{1}{2} m v^2 \]
Dividing through by \( m \) and simplifying:
\[ 10gl = 2gL + \frac{1}{2} v^2 \]
Rearranging gives:
\[ \frac{1}{2} v^2 = 10gl - 2gL \]
\[ v^2 = 20gl - 4gL \]
\[ v^2 = 4g(5l - L) \]

### Step 4: Substitute \( v^2 \) back into the centripetal force equation
Now we substitute \( v^2 \) into the centripetal force equation:
\[ T + mg = \frac{m(4g(5l - L))}{L} \]
This simplifies to:
\[ T = \frac{4mg(5l - L)}{L} - mg \]
Factoring out \( mg \):
\[ T = mg \left( \frac{4(5l - L)}{L} - 1 \right) \]

### Step 5: Find the minimum tension
To find the minimum tension, we evaluate \( T \) at the highest point where \( L = l \):
\[ T = mg \left( \frac{4(5l - l)}{l} - 1 \right) \]
\[ T = mg \left( \frac{4(4l)}{l} - 1 \right) \]
\[ T = mg(16 - 1) = 15mg \]

Thus, the minimum value of tension in the string is:
\[ \boxed{15mg} \]