A particle suspended by a thread of length T is projected horizontally with a velocity `sqrt(3gl)` at the lowest point. The height from the bottom at which the tension in the string becomes zero is:

Options

A. `(4l)/3`
B. `(2l)/3`
C. `(5l)/3`
D. `l/3`

Answer

To solve the problem, we will analyze the motion of the particle and apply the principles of energy conservation and circular motion.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle is suspended by a thread of length \( l \) and is projected horizontally with a velocity \( v = \sqrt{3gl} \) at the lowest point (point A).
- We need to find the height from the bottom (point A) at which the tension in the string becomes zero (point B).

2. **Identify Forces at Point B**:
- At point B, the tension in the string is zero. This means that the only force acting on the particle is its weight, which provides the necessary centripetal force for circular motion.
- The centripetal force required for the particle to move in a circular path is given by \( F_c = \frac{mv^2}{l} \).
- At point B, the gravitational force acting on the particle is \( F_g = mg \cos \theta \), where \( \theta \) is the angle between the vertical and the string at point B.

3. **Setting Up the Equation**:
- Since the tension is zero at point B, we can set the centripetal force equal to the component of the gravitational force:
\[ \frac{mv^2}{l} = mg \cos \theta \]
- Simplifying this gives:
\[ v^2 = gl \cos \theta \]

4. **Using Conservation of Energy**:
- The total mechanical energy at point A (the lowest point) is purely kinetic:
\[ E_A = \frac{1}{2} mv^2 = \frac{1}{2} m (3gl) = \frac{3mgl}{2} \]
- At point B, the energy is a combination of potential and kinetic energy:
\[ E_B = mgh + \frac{1}{2} mv^2 \]
- The height \( h \) at point B can be expressed as \( h = l(1 - \cos \theta) \).

5. **Equating Energies**:
- By conservation of energy, we have:
\[ E_A = E_B \]
- Substituting the expressions for \( E_A \) and \( E_B \):
\[ \frac{3mgl}{2} = mg(l(1 - \cos \theta)) + \frac{1}{2} mv^2 \]

6. **Substituting for \( v^2 \)**:
- From the centripetal force equation, we have \( v^2 = gl \cos \theta \). Substituting this into the energy equation gives:
\[ \frac{3mgl}{2} = mg(l(1 - \cos \theta)) + \frac{1}{2} m(gl \cos \theta) \]
- Simplifying this leads to:
\[ \frac{3}{2} = (1 - \cos \theta) + \frac{1}{2} \cos \theta \]
- Rearranging gives:
\[ \frac{3}{2} = 1 - \cos \theta + \frac{1}{2} \cos \theta \]
- This simplifies to:
\[ \frac{3}{2} = 1 - \frac{1}{2} \cos \theta \]
- Solving for \( \cos \theta \) yields:
\[ \cos \theta = \frac{1}{3} \]

7. **Finding the Height**:
- The height \( h \) from the lowest point (point A) to point B is given by:
\[ h = l(1 - \cos \theta) = l\left(1 - \frac{1}{3}\right) = l\left(\frac{2}{3}\right) = \frac{2l}{3} \]

### Final Answer:
The height from the bottom at which the tension in the string becomes zero is \( \frac{2l}{3} \).