A particle is released from rest from a tower of height 3h. The ratio of time intervals for fall of equal height h i.e. `t_(1):t_(2):t_(3)` is :

Options

A. `t_1 : t_2 : t_3 = 3 : 2 : 1`
B. `t_1 : t_2 : t_3 = 1 : (sqrt 2 - 1) : (sqrt 3 - 2)`
C. `t_1 : t_2 : t_3 = sqrt 3 : sqrt 2 : 1`
D. `t_1 : t_2 : t_3 = 1 : (sqrt 2 - 1) : (sqrt 3 - sqrt 2)`

Explanation

To solve the problem of finding the ratio of time intervals \( t_1 : t_2 : t_3 \) for a particle falling from a height of \( 3h \), we will use the equations of motion. The particle is released from rest, meaning its initial velocity \( u = 0 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The particle falls from a height of \( 3h \).
- We need to find the time intervals \( t_1 \), \( t_2 \), and \( t_3 \) for the particle to fall equal heights of \( h \).

2. **Using the Equation of Motion**:
- The equation of motion we will use is:
\[ s = ut + \frac{1}{2} g t^2 \]
- Since the particle is released from rest, \( u = 0 \), and the equation simplifies to:
\[ s = \frac{1}{2} g t^2 \]

3. **Finding \( t_1 \)**:
- For the first interval, the particle falls a height \( h \):
\[ h = \frac{1}{2} g t_1^2 \]
- Rearranging gives:
\[ t_1^2 = \frac{2h}{g} \implies t_1 = \sqrt{\frac{2h}{g}} \]

4. **Finding \( t_2 \)**:
- For the second interval, the particle falls from height \( h \) to height \( 2h \). The total distance fallen is \( 2h \):
\[ 2h = \frac{1}{2} g (t_1 + t_2)^2 \]
- We already found \( t_1 \), so substituting gives:
\[ 2h = \frac{1}{2} g \left(\sqrt{\frac{2h}{g}} + t_2\right)^2 \]
- Simplifying this expression will yield \( t_2 \).

5. **Finding \( t_3 \)**:
- For the third interval, the particle falls from height \( 2h \) to height \( 3h \). The total distance fallen is \( 3h \):
\[ 3h = \frac{1}{2} g (t_1 + t_2 + t_3)^2 \]
- Again, substituting the values of \( t_1 \) and \( t_2 \) will allow us to solve for \( t_3 \).

6. **Calculating the Ratios**:
- Now we have \( t_1 \), \( t_2 \), and \( t_3 \) in terms of \( h \) and \( g \).
- The ratios can be expressed as:
\[ t_1 : t_2 : t_3 = 1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2}) \]

### Final Answer:
The ratio of time intervals for the fall of equal height \( h \) is:
\[ t_1 : t_2 : t_3 = 1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2}) \]