A brass sphere of mass `0.2kg` falls freely from a height of `20m` and bounces to a height of `8m` from the ground. If the dissipated energy in this process is absorbed by the sphere the eise in its temperature is (specific heat of brass `=360 J kg^(-1)K^(-1)` , `g=10ms^(-2)`)

Options

A. `0.33^(@)C`
B. `0.66^(@)C`
C. `0.77^(@)C`
D. `0.88^(@)C`

Explanation

To solve the problem, we need to determine the rise in temperature of the brass sphere after it falls and bounces back. We will follow these steps:

### Step 1: Calculate the initial potential energy (PE_initial)
The initial height from which the sphere falls is \( h_1 = 20 \, m \). The potential energy at this height is given by the formula:

\[ PE_{\text{initial}} = mgh_1 \]

Where:
- \( m = 0.2 \, kg \) (mass of the sphere)
- \( g = 10 \, m/s^2 \) (acceleration due to gravity)
- \( h_1 = 20 \, m \)

Substituting the values:

\[ PE_{\text{initial}} = 0.2 \times 10 \times 20 = 40 \, J \]

### Step 2: Calculate the final potential energy (PE_final)
The height to which the sphere bounces back is \( h_2 = 8 \, m \). The potential energy at this height is:

\[ PE_{\text{final}} = mgh_2 \]

Substituting the values:

\[ PE_{\text{final}} = 0.2 \times 10 \times 8 = 16 \, J \]

### Step 3: Calculate the energy lost during the fall
The energy lost during the fall is the difference between the initial and final potential energies:

\[ \text{Energy lost} = PE_{\text{initial}} - PE_{\text{final}} \]

Substituting the values:

\[ \text{Energy lost} = 40 \, J - 16 \, J = 24 \, J \]

### Step 4: Relate the energy lost to the rise in temperature
The energy lost is absorbed by the sphere and causes a rise in temperature. The formula relating heat energy to temperature change is:

\[ Q = mc\Delta T \]

Where:
- \( Q \) is the heat energy absorbed (which is equal to the energy lost)
- \( m = 0.2 \, kg \) (mass of the sphere)
- \( c = 360 \, J/(kg \cdot K) \) (specific heat of brass)
- \( \Delta T \) is the rise in temperature

Rearranging the equation to solve for \( \Delta T \):

\[ \Delta T = \frac{Q}{mc} \]

Substituting the values:

\[ \Delta T = \frac{24 \, J}{0.2 \, kg \times 360 \, J/(kg \cdot K)} = \frac{24}{72} = \frac{1}{3} \, K \]

### Step 5: Convert the temperature rise to degrees Celsius
Since a change of 1 K is equivalent to a change of 1 °C, we have:

\[ \Delta T = \frac{1}{3} \, °C \approx 0.33 \, °C \]

### Final Answer
The rise in temperature of the brass sphere is approximately \( 0.33 \, °C \).

---