A block of ice with mass m falls into a lake. After impact, a mass of ice `(m)/(5)` melts. Both the block of ice and the lake have a temperature of `0^(@)C`. If L represents the latent heat of fusion, the distance the ice falls before striking the surface is

Options

A. A) `(L)/(5g)`
B. B) `(5L)/(g)`
C. C) `(gL)/(5m)`
D. D) `(mL)/(5g)`

Answer

To solve the problem, we need to analyze the energy transformations that occur when the block of ice falls into the lake and melts some of the ice upon impact.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the mass of the ice block be \( m \).
- The mass of ice that melts after impact is \( \frac{m}{5} \).
- Let \( h \) be the height from which the ice block falls.
- Let \( L \) be the latent heat of fusion.

2. **Calculate the Potential Energy (PE)**:
- When the ice block falls from height \( h \), its potential energy at the height is given by:
\[ PE = mgh \]
- Here, \( g \) is the acceleration due to gravity.

3. **Calculate the Heat Required to Melt Ice**:
- The amount of heat required to melt the mass of ice \( \frac{m}{5} \) is given by:
\[ Q = \frac{m}{5} L \]
- This heat is absorbed from the kinetic energy of the ice block upon impact.

4. **Equate Potential Energy and Heat**:
- According to the conservation of energy, the potential energy lost by the ice block is equal to the heat gained by the ice to melt:
\[ mgh = \frac{m}{5} L \]

5. **Simplify the Equation**:
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{L}{5} \]

6. **Solve for Height \( h \)**:
- Rearranging the equation to solve for \( h \):
\[ h = \frac{L}{5g} \]

### Final Result:
The distance the ice falls before striking the surface is:
\[ h = \frac{L}{5g} \]