Two identical masses of `5` kg each fall on a wheel from a height of `10` m . The wheel disturbs a mass of `2` kg water, the rise in temperature of water will be :-

Options

A. `2.6^(@)C`
B. `1.2^(@)C`
C. `0.32^(@)C`
D. `0.12^(@)C`

Explanation

To solve the problem, we need to calculate the rise in temperature of the water when two identical masses fall on a wheel from a height of 10 m. We will use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Calculate the Potential Energy of the Falling Masses**:
The potential energy (PE) of an object at height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the object (in kg)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height (in m)

For one mass of \( 5 \, \text{kg} \) falling from a height of \( 10 \, \text{m} \):
\[ PE_1 = 5 \times 9.81 \times 10 = 490.5 \, \text{J} \]
Since there are two identical masses:
\[ PE_{\text{total}} = 2 \times PE_1 = 2 \times 490.5 = 981 \, \text{J} \]

2. **Use the Energy to Heat the Water**:
The potential energy lost by the masses will be converted into heat energy gained by the water. The heat energy gained by the water can be expressed using the formula:
\[ Q = mc\Delta T \]
where:
- \( Q \) = heat energy (in J)
- \( m \) = mass of the water (in kg)
- \( c \) = specific heat capacity of water (approximately \( 4200 \, \text{J/kg°C} \))
- \( \Delta T \) = rise in temperature (in °C)

Rearranging the formula to find \( \Delta T \):
\[ \Delta T = \frac{Q}{mc} \]

3. **Substituting Values**:
Now, substituting the values we have:
- \( Q = 981 \, \text{J} \)
- \( m = 2 \, \text{kg} \)
- \( c = 4200 \, \text{J/kg°C} \)

\[ \Delta T = \frac{981}{2 \times 4200} \]
\[ \Delta T = \frac{981}{8400} \approx 0.1168 \, \text{°C} \]

4. **Final Result**:
The rise in temperature of the water is approximately \( 0.1168 \, \text{°C} \).