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 step by step, we will use the principles of energy conservation and heat transfer.

### Step 1: Calculate the potential energy of the falling masses
The potential energy (PE) of an object is given by the formula:
\[ \text{PE} = mgh \]
where:
- \( m \) is the mass (in kg),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( h \) is the height (in m).

Since there are two identical masses of \( 5 \, \text{kg} \) each falling from a height of \( 10 \, \text{m} \), we calculate the total potential energy as follows:
\[ \text{Total PE} = 2 \times m \times g \times h = 2 \times 5 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} \]
\[ \text{Total PE} = 2 \times 5 \times 10 \times 10 = 1000 \, \text{J} \]

### Step 2: Relate the potential energy to the heat gained by water
The work done (or energy transferred) by the falling masses will be converted into heat energy (Q) that raises the temperature of the water. The heat gained by the water can be expressed as:
\[ Q = ms\Delta T \]
where:
- \( m \) is the mass of the water (in kg),
- \( s \) is the specific heat capacity of water (approximately \( 4200 \, \text{J/(kg°C)} \)),
- \( \Delta T \) is the change in temperature (in °C).

### Step 3: Set the potential energy equal to the heat gained by the water
From the conservation of energy, we have:
\[ \text{Total PE} = Q \]
Thus,
\[ 1000 \, \text{J} = m \cdot s \cdot \Delta T \]

### Step 4: Substitute known values into the equation
We know:
- \( m = 2 \, \text{kg} \) (mass of water),
- \( s = 4200 \, \text{J/(kg°C)} \).

Substituting these values into the equation gives:
\[ 1000 = 2 \cdot 4200 \cdot \Delta T \]

### Step 5: Solve for \( \Delta T \)
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{1000}{2 \cdot 4200} \]
\[ \Delta T = \frac{1000}{8400} \]
\[ \Delta T = \frac{1}{8.4} \approx 0.119 \, \text{°C} \]

### Final Answer
Thus, the rise in temperature of the water will be approximately \( 0.12 \, \text{°C} \).

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