A particle is moving in a circular path in the vertical plane. It is attached at one end of a string of length `l` whose other end is fixed. The velocity at lowest point is `u` . The tension in the string `T` and acceleration of the particle is a at any position. Then `T.a` is zero at highest point if

Options

A. `ugtsqrt(5gl)`
B. `u=sqrt(5gl)`
C. Both (a) and (b) are correct
D. Both (a) and (b) are wrong

Explanation

To solve the problem step by step, we will analyze the motion of the particle in a vertical circular path and apply the principles of dynamics and energy conservation.

### Step 1: Understand the Forces Acting on the Particle
At the highest point of the circular path, the forces acting on the particle are:
- The gravitational force (mg) acting downward.
- The tension in the string (T) also acting downward.

At the highest point, the centripetal force required to keep the particle moving in a circle is provided by the combination of the gravitational force and the tension in the string. Therefore, we can write the equation for centripetal force as:
\[ T + mg = \frac{mv^2}{l} \]
where \( v \) is the velocity of the particle at the highest point and \( l \) is the length of the string (which is also the radius of the circular path).

### Step 2: Apply Conservation of Energy
To find the velocity \( v \) at the highest point, we can use the conservation of mechanical energy. The total mechanical energy at the lowest point (point A) is equal to the total mechanical energy at the highest point (point B).

At the lowest point, the energy is purely kinetic:
\[ E_A = \frac{1}{2}mu^2 \]

At the highest point, the energy is the sum of kinetic and potential energy:
\[ E_B = \frac{1}{2}mv^2 + mg(2l) \]
where \( 2l \) is the height gained when moving from the lowest point to the highest point.

Setting the energies equal gives us:
\[ \frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg(2l) \]

### Step 3: Solve for \( v^2 \)
Rearranging the equation to solve for \( v^2 \):
\[ \frac{1}{2}mu^2 - mg(2l) = \frac{1}{2}mv^2 \]
Multiplying through by 2 to eliminate the fraction:
\[ mu^2 - 4mgl = mv^2 \]
Dividing by \( m \):
\[ u^2 - 4gl = v^2 \]

### Step 4: Set Tension to Zero
According to the problem, we need to find the condition when \( T.a = 0 \) at the highest point. This implies that the tension \( T \) must be zero at the highest point:
\[ 0 + mg = \frac{mv^2}{l} \]
Substituting \( T = 0 \) into the centripetal force equation gives:
\[ mg = \frac{mv^2}{l} \]
Canceling \( m \) from both sides:
\[ g = \frac{v^2}{l} \]
Substituting \( v^2 = u^2 - 4gl \) into this equation:
\[ g = \frac{u^2 - 4gl}{l} \]

### Step 5: Solve for \( u \)
Rearranging gives:
\[ gl = u^2 - 4gl \]
Adding \( 4gl \) to both sides:
\[ 5gl = u^2 \]
Taking the square root:
\[ u = \sqrt{5gl} \]

### Conclusion
Thus, the initial velocity \( u \) at the lowest point must be \( \sqrt{5gl} \) for the tension in the string to be zero at the highest point.