A stationary objected at `0^(@)C` and weighing 3.5 kg falls from a height of 2000m on a snow mounation at `0^(@)C`. If the temperature of the object just before hitting the snow is `0^(@)C` and the object comes to rest immediately `(g = 10 m//s^(2))` and (latent heat of ice = `3.5 xx 10^(5) "joule"//s)` then the mass of ice that will melt is -

Options

A. 2 gm of ice
B. 20 gm of ice
C. 200 gm of ice
D. 2 kg of ice

Explanation

To solve the problem, we will use the principle of conservation of energy. The potential energy of the object when it is at a height will be converted into the latent heat required to melt the ice.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the object (m) = 3.5 kg
- Height (h) = 2000 m
- Acceleration due to gravity (g) = 10 m/s²
- Latent heat of ice (L) = 3.5 × 10^5 J/kg

2. **Calculate the Potential Energy (PE) of the Object:**
The potential energy (PE) of the object just before it hits the snow can be calculated using the formula:
\[ PE = m \cdot g \cdot h \]
Substituting the values:
\[ PE = 3.5 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 2000 \, \text{m} \]
\[ PE = 3.5 \cdot 10 \cdot 2000 = 70000 \, \text{J} \]

3. **Relate Potential Energy to the Latent Heat of Ice:**
The potential energy will be used to melt the ice. The relationship can be expressed as:
\[ PE = m' \cdot L \]
where \( m' \) is the mass of ice melted.

4. **Rearranging the Equation:**
We can rearrange the equation to find the mass of ice melted:
\[ m' = \frac{PE}{L} \]

5. **Substituting the Values:**
Now, substitute the values of PE and L:
\[ m' = \frac{70000 \, \text{J}}{3.5 \times 10^5 \, \text{J/kg}} \]
\[ m' = \frac{70000}{350000} = 0.2 \, \text{kg} \]

6. **Convert Mass to Grams:**
Since the question asks for the mass of ice in grams:
\[ m' = 0.2 \, \text{kg} = 0.2 \times 1000 \, \text{g} = 200 \, \text{g} \]

### Final Answer:
The mass of ice that will melt is **200 grams**.

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