Hailstone at `0^@C` from a height of 1 km on an insulating surface converting whole of its kinetic energy into heat. What part of it will melt? (`g=10 m//s`)

Options

A. `(1)/(33)`
B. `(1)/(8)`
C. `(1)/(33) xx 10^(-4)`
D. All of it will melt

Explanation

To solve the problem of how much of the hailstone will melt when it falls from a height of 1 km, we will follow these steps:

### Step 1: Calculate the Potential Energy (PE) of the Hailstone
The potential energy of the hailstone at a height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the hailstone (in kg)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( h \) = height (1000 m)

Substituting the values:
\[ PE = m \cdot 10 \cdot 1000 = 10000m \text{ J} \]

### Step 2: Convert Potential Energy to Kinetic Energy (KE)
As the hailstone falls, its potential energy converts into kinetic energy. At the moment just before it hits the surface, all the potential energy has been converted to kinetic energy:
\[ KE = PE = 10000m \text{ J} \]

### Step 3: Convert Kinetic Energy to Heat Energy
When the hailstone hits the insulating surface, all of its kinetic energy is converted into heat energy. This heat energy will be used to melt part of the hailstone:
\[ \text{Heat Energy} = KE = 10000m \text{ J} \]

### Step 4: Relate Heat Energy to the Mass of Ice Melted
The heat energy required to melt a mass \( k \) of ice is given by:
\[ Q = k \cdot L \]
where:
- \( L \) = latent heat of fusion of ice (approximately \( 330 \times 10^3 \) J/kg)

Setting the heat energy equal to the energy required to melt the ice:
\[ 10000m = k \cdot (330 \times 10^3) \]

### Step 5: Solve for \( k \)
Rearranging the equation to find \( k \):
\[ k = \frac{10000m}{330 \times 10^3} \]
\[ k = \frac{10000}{330000} m \]
\[ k = \frac{1}{33} m \]

### Step 6: Interpret the Result
The value \( k = \frac{1}{33} m \) indicates that \( \frac{1}{33} \) of the mass of the hailstone will melt. This means that if the mass of the hailstone is \( m \), then \( \frac{1}{33} \) of it will turn into water upon impact.

### Final Answer
Thus, the part of the hailstone that will melt is \( \frac{1}{33} \) of its total mass.

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