A tuning is dug along the diameter of the earth. There is particle of mass `m` at the centre of the tunel. Find the minimum velocity given to the particle so that is just reaches to the surface of the earth. `(R =` radius of earth)

Options

A. `sqrt((GM)/(R))`
B. `sqrt((GM)/(2R))`
C. `sqrt((2GM)/(R))`
D. it will reach with the help of negligible velocity

Explanation

To solve the problem of finding the minimum velocity given to a particle of mass \( m \) at the center of a tunnel dug along the diameter of the Earth, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understanding the Energy at the Center of the Earth
At the center of the Earth, the particle has only kinetic energy since the gravitational potential energy is at a minimum. The total mechanical energy (TME) at the center can be expressed as:
\[ \text{TME}_{\text{initial}} = \text{K.E.} + \text{P.E.} \]
At the center, the potential energy \( P.E. \) is given by:
\[ P.E. = -\frac{3GMm}{2R} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 2: Energy at the Surface of the Earth
When the particle reaches the surface of the Earth, its velocity becomes zero, and thus its kinetic energy is zero. The potential energy at the surface is given by:
\[ P.E. = -\frac{GMm}{R} \]
At this point, the total mechanical energy is:
\[ \text{TME}_{\text{final}} = 0 + \left(-\frac{GMm}{R}\right) = -\frac{GMm}{R} \]

### Step 3: Applying Conservation of Energy
According to the conservation of energy, the total mechanical energy at the center must equal the total mechanical energy at the surface:
\[ \text{TME}_{\text{initial}} = \text{TME}_{\text{final}} \]
Substituting the expressions we have:
\[ \frac{1}{2}mv^2 - \frac{3GMm}{2R} = -\frac{GMm}{R} \]

### Step 4: Simplifying the Equation
Now, we can rearrange the equation:
\[ \frac{1}{2}mv^2 = -\frac{GMm}{R} + \frac{3GMm}{2R} \]
Combining the terms on the right side:
\[ \frac{1}{2}mv^2 = \frac{3GMm}{2R} - \frac{2GMm}{2R} = \frac{GMm}{2R} \]

### Step 5: Solving for the Velocity
Now, we can solve for \( v \):
\[ \frac{1}{2}mv^2 = \frac{GMm}{2R} \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2}v^2 = \frac{GM}{2R} \]
Multiplying both sides by 2:
\[ v^2 = \frac{GM}{R} \]
Taking the square root:
\[ v = \sqrt{\frac{GM}{R}} \]

### Final Answer
Thus, the minimum velocity given to the particle so that it just reaches the surface of the Earth is:
\[ v = \sqrt{\frac{GM}{R}} \]