A container of uniform cross-section has a height of 14 m. Upto what height (in metre) water of refractive index `4//3` should be filled for normal viewing.

Explanation

To solve the problem of determining the height to which water should be filled in a container for normal viewing, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a container with a uniform cross-section and a total height of 14 meters.
- We need to find the height \( h \) of water that should be filled in the container, given that the refractive index of water is \( \frac{4}{3} \).

2. **Set Up the Relationship**:
- When viewing from the top of the container, the apparent depth of the water is less than the actual depth due to refraction.
- The relationship between the actual depth \( h \) and the apparent depth can be expressed as:
\[ \text{Apparent Depth} = \frac{h}{\mu} \]
- Here, \( \mu \) is the refractive index of water, which is \( \frac{4}{3} \).

3. **Determine the Apparent Depth**:
- The apparent depth when viewed from the top is given by:
\[ \text{Apparent Depth} = 14 - h \]
- This means that the apparent depth is the total height of the container minus the height of the water.

4. **Set Up the Equation**:
- From the relationships established, we can equate the two expressions for apparent depth:
\[ \frac{h}{\frac{4}{3}} = 14 - h \]

5. **Clear the Fraction**:
- To eliminate the fraction, multiply both sides of the equation by \( \frac{4}{3} \):
\[ h = \frac{4}{3}(14 - h) \]

6. **Distribute and Rearrange**:
- Distributing \( \frac{4}{3} \):
\[ h = \frac{56}{3} - \frac{4}{3}h \]
- Rearranging gives:
\[ h + \frac{4}{3}h = \frac{56}{3} \]
- This simplifies to:
\[ \frac{7}{3}h = \frac{56}{3} \]

7. **Solve for \( h \)**:
- Multiply both sides by \( \frac{3}{7} \):
\[ h = \frac{56}{7} = 8 \]

8. **Conclusion**:
- The height of water that should be filled in the container for normal viewing is \( h = 8 \) meters.

### Final Answer:
The height of water that should be filled in the container is **8 meters**.