The time of flight of a body becomes `n` times the original value if its speed is made `n` times.
This due to the range of the projectile which becomes `n` times.

Options

A. Statement (I) is true, Statement (II) is true , statement (II) is the correct explanation for Statement (I).
B. Statement (I) is true, Statement (II) is true , statement (II) is not the correct explanation for Statement (I).
C. Statement (I) is true, Statement (II) is false.
D. Statement (i) is false, Statement (II) is true.

Explanation

To solve the problem, we need to analyze the relationship between the time of flight, speed, and range of a projectile.

### Step-by-Step Solution:

1. **Understanding Time of Flight**:
The time of flight \( T \) of a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Effect of Increasing Speed**:
If the speed \( u \) is increased to \( nu \) (where \( n \) is a factor by which the speed is increased), the new time of flight \( T' \) becomes:
\[ T' = \frac{2(nu) \sin \theta}{g} = n \left(\frac{2u \sin \theta}{g}\right) = nT \]
This shows that the time of flight becomes \( n \) times the original time of flight.

3. **Understanding Range**:
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
If the speed \( u \) is increased to \( nu \), the new range \( R' \) becomes:
\[ R' = \frac{(nu)^2 \sin 2\theta}{g} = \frac{n^2 u^2 \sin 2\theta}{g} = n^2 R \]
This means that the range becomes \( n^2 \) times the original range.

4. **Conclusion**:
The statement that the time of flight becomes \( n \) times the original value when the speed is made \( n \) times is true. However, the statement that the range becomes \( n \) times is false; it actually becomes \( n^2 \) times the original range.

### Final Answer:
- Statement 1: True
- Statement 2: False