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)xx10^-4`
D. all of it

Explanation

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

### Step 1: Calculate the potential energy (PE) of the hailstone
The potential energy (PE) of an object at 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 (1 km = 1000 m)

Substituting the values:
\[ PE = m \cdot 10 \cdot 1000 = 10000m \text{ (in Joules)} \]

### Step 2: Relate potential energy to the heat required to melt part of the hailstone
When the hailstone hits the ground, all of its potential energy is converted into heat. The heat \( Q \) required to melt a mass \( k \cdot m \) of ice is given by:
\[ Q = k \cdot m \cdot L \]
where:
- \( L \) = latent heat of fusion of ice (approximately 80 kcal/kg)

To convert this to Joules, we use the conversion factor:
\[ 1 \text{ kcal} = 4184 \text{ Joules} \]
Thus,
\[ L = 80 \cdot 4184 = 334720 \text{ Joules/kg} \]

### Step 3: Set the potential energy equal to the heat required
Since all potential energy is converted into heat:
\[ 10000m = k \cdot m \cdot 334720 \]

### Step 4: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 10000 = k \cdot 334720 \]

### Step 5: Solve for \( k \)
Now, we can solve for \( k \):
\[ k = \frac{10000}{334720} \]

Calculating this gives:
\[ k \approx 0.0299 \]

### Step 6: Interpret the result
The value of \( k \) represents the fraction of the hailstone that will melt. Therefore, approximately \( 0.0299 \) or about \( 2.99\% \) of the hailstone will melt upon hitting the ground.

### Final Answer
Approximately \( 2.99\% \) of the hailstone will melt.

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