A body is allowed to fall from a height of `100 m`. If the time taken for the first `50 m` is `t_1` and for the remaining `50 m` is `t_2` then :

Options

A. `t_1 = t_2`
B. `t_1 gt t_2`
C. `t_1 lt t_2`
D. depends upon the mass.

Explanation

To solve the problem, we need to find the relationship between the time taken to fall the first 50 meters (T1) and the time taken to fall the next 50 meters (T2) when a body falls from a height of 100 meters.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- The body falls from a height of 100 meters.
- The distance is divided into two parts: the first 50 meters and the second 50 meters.
- We need to find the relationship between T1 (time for the first 50 m) and T2 (time for the second 50 m).

2. **Using the Equation of Motion:**
- For free fall, the equation of motion is given by:
\[ S = ut + \frac{1}{2}gt^2 \]
- Here, \( S \) is the distance fallen, \( u \) is the initial velocity (which is 0 for free fall), \( g \) is the acceleration due to gravity, and \( t \) is the time taken.

3. **Calculating T1:**
- For the first 50 meters:
\[ 50 = 0 \cdot T1 + \frac{1}{2}gT1^2 \]
- Simplifying this gives:
\[ 50 = \frac{1}{2}gT1^2 \implies T1^2 = \frac{100}{g} \implies T1 = \sqrt{\frac{100}{g}} = \frac{10}{\sqrt{g}} \]

4. **Calculating Total Time for 100 Meters:**
- For the total distance of 100 meters:
\[ 100 = 0 \cdot T + \frac{1}{2}gT^2 \]
- This simplifies to:
\[ 100 = \frac{1}{2}gT^2 \implies T^2 = \frac{200}{g} \implies T = \sqrt{\frac{200}{g}} = \frac{10\sqrt{2}}{\sqrt{g}} \]

5. **Finding T2:**
- Since \( T \) is the total time for the fall, we can express \( T2 \) as:
\[ T2 = T - T1 \]
- Substituting the values we calculated:
\[ T2 = \frac{10\sqrt{2}}{\sqrt{g}} - \frac{10}{\sqrt{g}} = \frac{10(\sqrt{2} - 1)}{\sqrt{g}} \]

6. **Finding the Relationship between T1 and T2:**
- Now we have:
\[ T1 = \frac{10}{\sqrt{g}} \quad \text{and} \quad T2 = \frac{10(\sqrt{2} - 1)}{\sqrt{g}} \]
- To find the relationship, we can compare \( T1 \) and \( T2 \):
\[ \frac{T2}{T1} = \frac{10(\sqrt{2} - 1)}{10} = \sqrt{2} - 1 \]
- Since \( \sqrt{2} \approx 1.414 \), we find:
\[ \sqrt{2} - 1 \approx 0.414 < 1 \]
- Therefore, \( T2 < T1 \).

### Conclusion:
Thus, we conclude that \( T1 > T2 \).