A body of mass m hangs at one end of a string of length l , the other end of which is fixed. It is given a horizontal velocity so that the string would just reach where it makes an angle of `60^(@)` with the vertical. The tension in the string at mean position is

Options

A. `2mg`
B. `mg`
C. `3mg`
D. `sqrt(3) mg`

Answer

To find the tension in the string at the mean position when a body of mass \( m \) is hanging from a string of length \( l \) and is given a horizontal velocity, we can follow these steps:

### Step 1: Understanding the System
The body is hanging from a fixed point and is given a horizontal velocity. As it swings, it will reach a position where the string makes an angle of \( 60^\circ \) with the vertical.

### Step 2: Determine the Velocity at the Mean Position
Using the work-energy theorem, we can find the velocity \( v \) of the mass at the mean position (the lowest point of the swing). The change in potential energy as the mass moves from the mean position to the height where it makes a \( 60^\circ \) angle can be expressed as:

\[ v = \sqrt{2g l (1 - \cos \theta)} \]

Here, \( \theta = 60^\circ \) and \( \cos 60^\circ = \frac{1}{2} \). Thus, we have:

\[ v = \sqrt{2g l (1 - \frac{1}{2})} = \sqrt{2g l \cdot \frac{1}{2}} = \sqrt{gl} \]

### Step 3: Apply the Forces at the Mean Position
At the mean position, the forces acting on the mass are:
- The tension \( T \) in the string acting upwards.
- The weight of the mass \( mg \) acting downwards.

According to the centripetal force requirement, the net force towards the center must equal the centripetal force:

\[ T - mg = \frac{mv^2}{l} \]

### Step 4: Substitute the Velocity
Substituting \( v^2 = gl \) into the equation gives:

\[ T - mg = \frac{m(gl)}{l} \]

This simplifies to:

\[ T - mg = mg \]

### Step 5: Solve for Tension
Now, we can solve for \( T \):

\[ T = mg + mg = 2mg \]

### Conclusion
Thus, the tension in the string at the mean position is:

\[ \boxed{2mg} \]