A coin is dropped in a lift. It takes time `t_(1)` to reach the floor when lift is stationary. It takes time `t_(2)` when lift is moving up with costant acceleration. Then

Options

A. `t_(1) = t_(2)`
B. `t_(1) gt t_(2)`
C. `t_(2) gt t_(1)`
D. `t_(1) gt gt t_(2)`

Explanation

To solve the problem, we need to analyze the motion of the coin in two different scenarios: when the lift is stationary and when the lift is moving upwards with constant acceleration.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- When the lift is stationary, the only force acting on the coin is gravity.
- When the lift is moving upwards with constant acceleration, the effective acceleration acting on the coin changes.

2. **Time Calculation when Lift is Stationary**:
- Let the height from which the coin is dropped be \( h \).
- The time taken \( t_1 \) for the coin to reach the floor when the lift is stationary can be derived from the equation of motion:
\[ h = \frac{1}{2} g t_1^2 \]
- Rearranging this gives:
\[ t_1 = \sqrt{\frac{2h}{g}} \]

3. **Time Calculation when Lift is Moving Upwards**:
- When the lift is moving upwards with a constant acceleration \( a \), the effective acceleration acting on the coin becomes \( g + a \) (since the coin is falling downwards while the lift is accelerating upwards).
- The time taken \( t_2 \) for the coin to reach the floor in this case can be calculated using the same equation of motion:
\[ h = \frac{1}{2} (g + a) t_2^2 \]
- Rearranging this gives:
\[ t_2 = \sqrt{\frac{2h}{g + a}} \]

4. **Comparing \( t_1 \) and \( t_2 \)**:
- We have:
\[ t_1 = \sqrt{\frac{2h}{g}} \quad \text{and} \quad t_2 = \sqrt{\frac{2h}{g + a}} \]
- To compare \( t_1 \) and \( t_2 \), we can square both sides:
\[ t_1^2 = \frac{2h}{g} \quad \text{and} \quad t_2^2 = \frac{2h}{g + a} \]
- Since \( g + a > g \), it follows that:
\[ \frac{2h}{g + a} < \frac{2h}{g} \]
- Therefore, we conclude that:
\[ t_2^2 < t_1^2 \implies t_2 < t_1 \]

5. **Conclusion**:
- Thus, the time taken for the coin to reach the floor when the lift is moving upwards with constant acceleration is less than the time taken when the lift is stationary:
\[ t_1 > t_2 \]

### Final Answer:
The correct option is that \( t_1 \) is greater than \( t_2 \).

---