A hole is drilled from the surface of earth to its centre. A particle is dropped from rest in the surface of earth in terms of its escape velocity on the surface of earth `v_(e)` is :

Options

A. `(v_(e))/(2)`
B. `v_(e)`
C. `sqrt(2)v_(e)`
D. `(v_(e))/(sqrt(2))`

Answer

To solve the problem, we need to analyze the motion of a particle dropped from the surface of the Earth to its center, and express its velocity at the center in terms of the escape velocity from the surface of the Earth.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( v_e \) from the surface of the Earth is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Initial Conditions**:
When the particle is dropped from the surface of the Earth, its initial velocity \( u \) is 0 (since it is dropped from rest). The potential energy \( U \) at the surface is:
\[ U_i = -\frac{GMm}{R} \]
where \( m \) is the mass of the particle.

3. **Final Conditions at the Center**:
At the center of the Earth, the potential energy \( U_f \) can be calculated using the formula for gravitational potential energy inside a uniform sphere:
\[ U_f = -\frac{3GMm}{2R} \]
The kinetic energy \( K_f \) at the center when the particle reaches its maximum speed \( v \) is:
\[ K_f = \frac{1}{2}mv^2 \]

4. **Applying Conservation of Energy**:
According to the conservation of mechanical energy:
\[ U_i + K_i = U_f + K_f \]
Since the initial kinetic energy \( K_i \) is 0, we have:
\[ -\frac{GMm}{R} = -\frac{3GMm}{2R} + \frac{1}{2}mv^2 \]

5. **Simplifying the Equation**:
Canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ -\frac{GM}{R} = -\frac{3GM}{2R} + \frac{1}{2}v^2 \]
Rearranging gives:
\[ \frac{1}{2}v^2 = -\frac{GM}{R} + \frac{3GM}{2R} \]
\[ \frac{1}{2}v^2 = \frac{GM}{2R} \]

6. **Finding the Velocity**:
Multiplying both sides by 2:
\[ v^2 = \frac{GM}{R} \]
Taking the square root:
\[ v = \sqrt{\frac{GM}{R}} \]

7. **Relating to Escape Velocity**:
We know that:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
Therefore:
\[ v = \sqrt{\frac{GM}{R}} = \frac{v_e}{\sqrt{2}} \]

### Final Answer:
The velocity of the particle when it reaches the center of the Earth is:
\[ v = \frac{v_e}{\sqrt{2}} \]