An iron ball and a wooden ball of the same radius are released from a height ‘ h ’ in vacuum. The time taken by both of them to reach the ground is

Options

A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-I
B. Statement-I is true, Statement-II is true, Statement-II is NOT a 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 of determining the time taken by an iron ball and a wooden ball of the same radius released from a height \( h \) in a vacuum, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- We have two balls: an iron ball and a wooden ball, both of the same radius.
- They are released from the same height \( h \) in a vacuum, meaning there is no air resistance.

2. **Identifying Forces**:
- In a vacuum, the only force acting on both balls is gravity.
- The acceleration due to gravity \( g \) is the same for both balls, regardless of their mass.

3. **Using the Equations of Motion**:
- For an object in free fall, the second equation of motion can be used:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s \) is the distance fallen (which is \( h \)), \( u \) is the initial velocity (which is 0 since they are released), \( a \) is the acceleration (which is \( g \)), and \( t \) is the time taken.

4. **Substituting Values**:
- Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ h = 0 + \frac{1}{2} g t^2 \]
- This can be rewritten as:
\[ h = \frac{1}{2} g t^2 \]

5. **Solving for Time \( t \)**:
- Rearranging the equation to solve for \( t \):
\[ t^2 = \frac{2h}{g} \]
- Taking the square root gives:
\[ t = \sqrt{\frac{2h}{g}} \]

6. **Conclusion**:
- Since both balls experience the same gravitational acceleration and are released from the same height, the time taken by both the iron ball and the wooden ball to reach the ground is the same:
\[ t = \sqrt{\frac{2h}{g}} \]