A body is thrown with a velocity equal to n times the escape velocity `(v_(e))`. Velocity of the body at large distance away will be :-

Options

A. `v_(e)sqrt(n^(2)-1)`
B. `v_(e)sqrt(n^(2)+1)`
C. `v_(e)sqrt(1-n^(2))`
D. None of these

Answer

To solve the problem, we need to find the velocity of a body thrown with a velocity equal to \( n \) times the escape velocity \( (v_e) \) at a large distance from the Earth.

### Step-by-Step Solution:

1. **Understand 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 Velocity**:
The body is thrown with an initial velocity \( v_0 = n v_e \).

3. **Mechanical Energy Conservation**:
The total mechanical energy (kinetic + potential) is conserved. At the surface of the Earth, the total mechanical energy \( E_i \) is:
\[ E_i = \text{Kinetic Energy} + \text{Potential Energy} \]
The initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2} m (n v_e)^2 = \frac{1}{2} m n^2 v_e^2 \]
The initial potential energy \( PE_i \) at the surface of the Earth is:
\[ PE_i = -\frac{GMm}{R} \]
Therefore, the total initial energy is:
\[ E_i = \frac{1}{2} m n^2 v_e^2 - \frac{GMm}{R} \]

4. **Final Energy at Large Distance**:
At a large distance, the potential energy \( PE_f \) is zero, and the total energy \( E_f \) is just the kinetic energy:
\[ E_f = KE_f = \frac{1}{2} m v^2 \]

5. **Setting Initial Energy Equal to Final Energy**:
Since mechanical energy is conserved, we have:
\[ E_i = E_f \]
Substituting the expressions for \( E_i \) and \( E_f \):
\[ \frac{1}{2} m n^2 v_e^2 - \frac{GMm}{R} = \frac{1}{2} m v^2 \]

6. **Simplifying the Equation**:
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} n^2 v_e^2 - \frac{GM}{R} = \frac{1}{2} v^2 \]

7. **Substituting \( v_e \)**:
Substitute \( v_e = \sqrt{\frac{2GM}{R}} \):
\[ \frac{1}{2} n^2 \left(\frac{2GM}{R}\right) - \frac{GM}{R} = \frac{1}{2} v^2 \]
Simplifying gives:
\[ \frac{n^2 GM}{R} - \frac{GM}{R} = \frac{1}{2} v^2 \]
\[ \frac{(n^2 - 1) GM}{R} = \frac{1}{2} v^2 \]

8. **Solving for \( v^2 \)**:
Multiply both sides by 2:
\[ \frac{2(n^2 - 1) GM}{R} = v^2 \]

9. **Taking the Square Root**:
Finally, taking the square root gives:
\[ v = \sqrt{\frac{2(n^2 - 1) GM}{R}} \]

10. **Substituting Back for \( v_e \)**:
Since \( \frac{2GM}{R} = v_e^2 \):
\[ v = v_e \sqrt{n^2 - 1} \]

### Final Answer:
The velocity of the body at a large distance away will be:
\[ v = v_e \sqrt{n^2 - 1} \]