A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel `=460 K//kg^(@)//C,g=10 m//s^(2)`)

Options

A. `0.01^@C`
B. `0.1^@C`
C. `1^@C`
D. `1.1^@C`

Explanation

To solve the problem step by step, we will follow the outlined process to find the rise in temperature of the steel ball after it bounces.

### Step 1: Calculate the initial and final potential energy
The potential energy (PE) of an object is given by the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the object (0.1 kg)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( h \) = height (in meters)

**Initial Potential Energy (PE_initial) at height 10 m:**
\[ PE_{initial} = mgh_i = 0.1 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 10 \, \text{J} \]

**Final Potential Energy (PE_final) at height 5.4 m:**
\[ PE_{final} = mgh_f = 0.1 \, \text{kg} \times 10 \, \text{m/s}^2 \times 5.4 \, \text{m} = 5.4 \, \text{J} \]

### Step 2: Calculate the loss in potential energy
The loss in potential energy (which is the energy dissipated) can be calculated as:
\[ \text{Loss in PE} = PE_{initial} - PE_{final} = 10 \, \text{J} - 5.4 \, \text{J} = 4.6 \, \text{J} \]

### Step 3: Relate the dissipated energy to heat absorbed by the ball
The energy lost by the ball is converted into heat, which raises the temperature of the ball. The heat absorbed (Q) can be expressed as:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( m \) = mass of the ball (0.1 kg)
- \( s \) = specific heat of steel (460 J/kg°C)
- \( \Delta T \) = rise in temperature (in °C)

### Step 4: Set the loss in potential energy equal to the heat absorbed
Since the dissipated energy is equal to the heat absorbed by the ball:
\[ 4.6 \, \text{J} = 0.1 \, \text{kg} \times 460 \, \text{J/kg°C} \times \Delta T \]

### Step 5: Solve for the rise in temperature (\( \Delta T \))
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{4.6 \, \text{J}}{0.1 \, \text{kg} \times 460 \, \text{J/kg°C}} = \frac{4.6}{46} = 0.1 \, °C \]

### Final Answer
The rise in temperature of the steel ball is:
\[ \Delta T = 0.1 \, °C \]

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