A boat having a length of 3 m and breadth of 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the mas is:

Options

A. 60 kg
B. 72 kg
C. 52 kg
D. 65 kg

Explanation

To solve the problem, we need to determine the mass of the man who causes the boat to sink by 1 cm when he gets on it. We will use the principles of buoyancy and Archimedes' principle.

### Step-by-Step Solution:

1. **Determine the dimensions of the boat:**
- Length (L) = 3 m
- Breadth (B) = 2 m
- The boat sinks by a height (h) = 1 cm = 0.01 m.

2. **Calculate the volume of water displaced by the boat when it sinks:**
- The volume (V) of water displaced can be calculated using the formula:
\[ V = L \times B \times h \]
- Substituting the values:
\[ V = 3 \, \text{m} \times 2 \, \text{m} \times 0.01 \, \text{m} = 0.06 \, \text{m}^3 \]

3. **Calculate the weight of the water displaced:**
- The weight of the water displaced is equal to the buoyant force acting on the boat, which is given by:
\[ \text{Weight of water} = \text{Volume} \times \text{Density} \times g \]
- The density of water (ρ) is approximately \(1000 \, \text{kg/m}^3\) and \(g\) (acceleration due to gravity) is approximately \(10 \, \text{m/s}^2\).
- Therefore:
\[ \text{Weight of water} = 0.06 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \]
- Calculating this gives:
\[ \text{Weight of water} = 0.06 \times 1000 \times 10 = 600 \, \text{N} \]

4. **Relate the weight of the man to the buoyant force:**
- According to Archimedes' principle, the weight of the man (W) is equal to the buoyant force (which is equal to the weight of the water displaced):
\[ W = \text{Weight of water displaced} = 600 \, \text{N} \]

5. **Calculate the mass of the man:**
- The weight of the man can also be expressed as:
\[ W = m \cdot g \]
- Rearranging gives:
\[ m = \frac{W}{g} \]
- Substituting the values:
\[ m = \frac{600 \, \text{N}}{10 \, \text{m/s}^2} = 60 \, \text{kg} \]

### Final Answer:
The mass of the man is **60 kg**.