Question:

In the container given below of dimensions (40 cm X 20 cm X 20 cm), four objects are dipped in water. Objects P and Q are made of some light material while objects R and S are made of iron and copper, respectively. The object P is a cube of edge 4 cm with 1/10th floating above water; object Q displaces 50 cc of water while floating. The volumes of objects R and S are 295.4 cc and 397 cc, respectively. If all the objects are removed from the container, what would be the new water level inside the container measured from the bottom?
In the container given below of dimensions (40 cm X 20 cm X 20 cm), four objects are dipped in water.

Updated On: Sep 8, 2025
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Correct Answer: 9

Solution and Explanation

To determine the new water level when all objects are removed, follow these steps:
1. **Understand the Container Volume**: The container dimensions are 40 cm x 20 cm x 20 cm.
2. **Calculate the Total Container Volume**:
\[ \text{Volume} = 40 \, \text{cm} \times 20 \, \text{cm} \times 20 \, \text{cm} = 16000 \, \text{cm}^3 \]
3. **Calculate the Immersed Volume of Each Object**:
- **Object P**:
- P is a cube with an edge of 4 cm:
\[ \text{Cube Volume} = 4^3 = 64 \, \text{cm}^3 \]
- Since 1/10th is above water:
\[ \text{Immersed Volume (P)} = \left( \frac{9}{10} \right) \times 64 = 57.6 \, \text{cm}^3 \]
- **Object Q**:
- Displaces 50 cc of water when floating.
\[ \text{Immersed Volume (Q)} = 50 \, \text{cm}^3 \]
- **Object R**:
- Volume of 295.4 cc, fully immersed.
\[ \text{Immersed Volume (R)} = 295.4 \, \text{cm}^3 \]
- **Object S**:
- Volume of 397 cc, fully immersed.
\[ \text{Immersed Volume (S)} = 397 \, \text{cm}^3 \]
4. **Calculate Total Water Displacement**:
\[ \text{Total Displacement} = 57.6 + 50 + 295.4 + 397 = 800 \, \text{cm}^3 \]
5. **Initial Water Volume**:
- Assume initial water filled the container at the level of total displacement:
\[ \text{Initial Water Volume} = 800 \, \text{cm}^3 \]
6. **New Water Level**:
- With objects removed, only the initial water remains:
- Cross-section of the container = 40 cm x 20 cm = 800 cm2.
- New height = Volume / Area:
\[ \text{New Water Height} = \frac{800 \, \text{cm}^3}{800 \, \text{cm}^2} = 1 \, \text{cm} \]
Since 1 cm is not within the specified range (9,9), there might be a discrepancy in assumptions; however, based on calculations with the given inputs, the new water level is 1 cm.
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