A particle is dropped from rest from a large height Assume `g` to be constant throughout the motion. The time taken by it to fall through successive distance of `1 m` each will be :

Options

A. All equal, being equal to `sqrt(2//g) second `
B. In the ratio of the square roots roots of the integers `1, 2, 3,.............
C. In the ratio of the disfference in the square roots of the integers, i.e., sqrt1, (sqrt2, -sqrt1),(sqrt3-sqrt2), (sqrt4-sqrt3)`......
D. In the ratio of the rectiprocals of the square roots of the integers, i.e., (1)/(sqrt1), (1)/(sqrt2), (1)/(sqrt(3)`,........

Explanation

To solve the problem of finding the time taken by a particle to fall through successive distances of 1 meter each when dropped from rest, we will use the equations of motion under constant acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( g \) be the acceleration due to gravity (constant).
- Let \( S_n \) be the distance fallen after \( n \) seconds.
- Let \( T_n \) be the time taken to fall through \( S_n \).

2. **Use the Equation of Motion**:
The equation of motion for an object falling from rest is given by:
\[ S = ut + \frac{1}{2}gt^2 \]
Since the particle is dropped from rest, the initial velocity \( u = 0 \). Therefore, the equation simplifies to:
\[ S_n = \frac{1}{2}gT_n^2 \]

3. **Calculate Distances**:
- For the first distance \( S_n \):
\[ S_n = \frac{1}{2}gT_n^2 \]
- For the distance \( S_{n+1} \) (which is \( S_n + 1 \)):
\[ S_{n+1} = \frac{1}{2}gT_{n+1}^2 \]

4. **Set Up the Equations**:
From the above, we have:
\[ S_n = \frac{1}{2}gT_n^2 \]
\[ S_{n+1} = S_n + 1 = \frac{1}{2}gT_{n+1}^2 \]

5. **Substituting for \( S_{n+1} \)**:
Substitute \( S_n \) into the equation for \( S_{n+1} \):
\[ \frac{1}{2}gT_n^2 + 1 = \frac{1}{2}gT_{n+1}^2 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{1}{2}gT_{n+1}^2 - \frac{1}{2}gT_n^2 = 1 \]
Factor out \( \frac{1}{2}g \):
\[ \frac{1}{2}g(T_{n+1}^2 - T_n^2) = 1 \]

7. **Solving for Time Difference**:
Multiply both sides by \( \frac{2}{g} \):
\[ T_{n+1}^2 - T_n^2 = \frac{2}{g} \]
This can be factored as:
\[ (T_{n+1} - T_n)(T_{n+1} + T_n) = \frac{2}{g} \]

8. **Finding the Time for 1 Meter**:
Let \( \Delta T = T_{n+1} - T_n \). Then:
\[ \Delta T (T_{n+1} + T_n) = \frac{2}{g} \]
As \( n \) becomes large, \( T_n \) and \( T_{n+1} \) approach the same value, so we can approximate:
\[ \Delta T \approx \frac{2}{g} \cdot \frac{1}{2T_n} = \frac{1}{gT_n} \]

9. **Final Expression**:
Thus, the time taken to fall through successive distances of 1 meter becomes:
\[ \Delta T \approx \sqrt{\frac{2}{g}} \cdot ( \sqrt{n + 1} - \sqrt{n} ) \]

### Conclusion:
The time taken by the particle to fall through successive distances of 1 meter is given by the expression derived above, which can be simplified further based on the values of \( n \).