A tank full of water has a small hole at its bottom. Let `t_(1)` be the time taken to empty first one third of the tank and `t_(2)` be the time taken to empty second one third of the tank and `t_(3)` be the time taken to empty rest of the tank then
(a). `t_(1)=t_(2)=t_(3)`
(b). `t_(1)gtt_(2)gtt_(3)`
(c). `t_(1)ltt_(2)ltt_(3)`
(d). `t_(1)gtt_(2)ltt_(3)`

Explanation

To solve the problem of a tank with a small hole at its bottom, we need to analyze the time taken to empty each segment of the tank.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a tank filled with water and a small hole at the bottom.
- We need to find the time taken to empty the first one-third of the tank (`t1`), the second one-third (`t2`), and the last one-third (`t3`).

2. **Applying Torricelli’s Law**:
- According to Torricelli's law, the speed of efflux of a fluid under the force of gravity through an orifice is given by \( v = \sqrt{2gh} \), where \( h \) is the height of the fluid above the hole.
- As the water level decreases, the height \( h \) decreases, which means the speed of water flowing out also decreases.

3. **Time to Empty Each Segment**:
- The time taken to empty the first one-third of the tank (`t1`) will be less than the time taken to empty the second one-third (`t2`), and `t2` will be less than the time taken to empty the last one-third (`t3`).
- This is because the water flows faster when the height is greater (during the first one-third) and slows down as the height decreases.

4. **Establishing the Relationship**:
- Since the flow rate decreases as the height decreases, we can conclude that:
\[ t1 < t2 < t3 \]
- Therefore, the correct relationship is:
\[ t1 < t2 < t3 \]

5. **Conclusion**:
- Based on the analysis, the correct answer is option (c): \( t1 < t2 < t3 \).