A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be `75^(@)C` . T is given by : (Given : room temperature = `30^(@)C`, specific heat of copper `=0.1 cal//gm^(@)C`)

Options

A. `0.63 xx 10^(-3) "rad/s"`
B. `0.28 xx 10^(-3) "rad/s"`
C. `1.1 xx 10^(-3) "rad/s"`
D. `0.83 xx 10^(-3) "rad/s"`

Answer

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the copper ball will be equal to the heat gained by the copper calorimeter and the water.

### Step-by-step Solution:

1. **Identify Given Values:**
- Mass of copper ball, \( m_1 = 100 \, \text{gm} \)
- Mass of copper calorimeter, \( m_2 = 100 \, \text{gm} \)
- Mass of water, \( m_3 = 170 \, \text{gm} \)
- Initial temperature of water and calorimeter, \( T_{initial} = 30^\circ C \)
- Final temperature of the system, \( T_{final} = 75^\circ C \)
- Specific heat of copper, \( c_{Cu} = 0.1 \, \text{cal/gm}^\circ C \)
- Specific heat of water, \( c_{water} = 1 \, \text{cal/gm}^\circ C \)

2. **Heat Lost by the Copper Ball:**
The heat lost by the copper ball when it cools down from temperature \( T \) to \( 75^\circ C \) is given by:
\[ Q_{lost} = m_1 \cdot c_{Cu} \cdot (T - T_{final}) = 100 \cdot 0.1 \cdot (T - 75) \]
Simplifying this, we get:
\[ Q_{lost} = 10(T - 75) \]

3. **Heat Gained by the Copper Calorimeter:**
The heat gained by the copper calorimeter as it warms up from \( 30^\circ C \) to \( 75^\circ C \) is:
\[ Q_{gained, calorimeter} = m_2 \cdot c_{Cu} \cdot (T_{final} - T_{initial}) = 100 \cdot 0.1 \cdot (75 - 30) \]
Simplifying this, we get:
\[ Q_{gained, calorimeter} = 10 \cdot 45 = 450 \, \text{cal} \]

4. **Heat Gained by the Water:**
The heat gained by the water as it warms up from \( 30^\circ C \) to \( 75^\circ C \) is:
\[ Q_{gained, water} = m_3 \cdot c_{water} \cdot (T_{final} - T_{initial}) = 170 \cdot 1 \cdot (75 - 30) \]
Simplifying this, we get:
\[ Q_{gained, water} = 170 \cdot 45 = 7650 \, \text{cal} \]

5. **Setting Up the Equation:**
According to the conservation of energy:
\[ Q_{lost} = Q_{gained, calorimeter} + Q_{gained, water} \]
Substituting the values we calculated:
\[ 10(T - 75) = 450 + 7650 \]
\[ 10(T - 75) = 8100 \]

6. **Solving for T:**
Dividing both sides by 10:
\[ T - 75 = 810 \]
\[ T = 810 + 75 = 885^\circ C \]

### Final Answer:
The temperature \( T \) of the copper ball is \( 885^\circ C \).