A large block of ice 10 cm thick with a vertical hole drilled through it is floating in a lake. The minimum length of the rope required to scoop out a bucket full of water through the hole is `(0.2 x)m`. Find the value of `x`. Take density of ice `=0.9" g/cm"^(3)`

Answer

To solve the problem, we need to analyze the situation of the block of ice floating in water and determine the minimum length of rope required to scoop out a bucket of water through the hole in the ice.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a block of ice that is 10 cm thick, floating on a lake.
- There is a vertical hole drilled through the ice.
- The density of ice is given as \(0.9 \, \text{g/cm}^3\).
- We need to find the minimum length of rope required to scoop out a bucket of water through the hole, which is given as \(0.2x\) meters.

2. **Establish the Forces**:
- The block of ice is floating, which means the weight of the ice is balanced by the buoyant force acting on it.
- Let \(A\) be the cross-sectional area of the ice block, and \(y\) be the depth of the water above the hole.

3. **Apply the Principle of Buoyancy**:
- The weight of the ice block can be expressed as:
\[ \text{Weight of ice} = V_{\text{ice}} \cdot \rho_{\text{ice}} \cdot g = A \cdot 10 \, \text{cm} \cdot 0.9 \, \text{g/cm}^3 \cdot g \]
- The buoyant force can be expressed as:
\[ \text{Buoyant force} = V_{\text{displaced}} \cdot \rho_{\text{water}} \cdot g = A \cdot (10 - y) \cdot 1 \, \text{g/cm}^3 \cdot g \]
- Setting the weight of the ice equal to the buoyant force gives:
\[ A \cdot 10 \cdot 0.9 \cdot g = A \cdot (10 - y) \cdot 1 \cdot g \]

4. **Canceling Common Terms**:
- Since \(A\) and \(g\) are common in both sides, we can cancel them out:
\[ 10 \cdot 0.9 = 10 - y \]

5. **Solving for \(y\)**:
- Rearranging the equation gives:
\[ 9 = 10 - y \implies y = 10 - 9 = 1 \, \text{cm} \]

6. **Relating \(y\) to the Length of the Rope**:
- The minimum length of the rope required to scoop out the bucket of water through the hole is given as \(0.2x\) meters.
- Since \(y\) is the depth of the water above the hole, we can equate:
\[ 0.2x = 1 \, \text{cm} \]
- Converting \(1 \, \text{cm}\) to meters gives \(0.01 \, \text{m}\):
\[ 0.2x = 0.01 \]

7. **Solving for \(x\)**:
- Dividing both sides by \(0.2\):
\[ x = \frac{0.01}{0.2} = 0.05 \, \text{m} \]
- Converting \(0.05 \, \text{m}\) to centimeters gives \(5 \, \text{cm}\).

### Final Answer:
The value of \(x\) is \(5\).