A 1.5 kg steel sphere will not fit through a circular hole in a 0.85 kg aluminium plate, because the radius of the sphere is 0.10% larger than the radius of the hole. If both the sphere and the plate are always kept at the same temperature, how much heat must be put into the two so the ball just passes through the hole ?

Options

A. `6.2 xx 10^(4)J`
B. `1.3 xx 10^(5)J`
C. `7.0 xx 10^(4)J`
D. `2.4 xx 10^(5)J`

Answer

To solve the problem, we need to determine how much heat must be added to both the steel sphere and the aluminum plate so that the sphere can pass through the hole in the plate. The key concept here is thermal expansion, which states that materials expand when heated.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the steel sphere, \( m_s = 1.5 \, \text{kg} \)
- Mass of the aluminum plate, \( m_a = 0.85 \, \text{kg} \)
- Radius of the sphere is 0.10% larger than the radius of the hole.

2. **Calculate the Difference in Radius:**
- Let the radius of the hole be \( r_h \).
- The radius of the sphere \( r_s = r_h \times (1 + 0.001) = 1.001 r_h \).
- The difference in radius \( \Delta r = r_s - r_h = 0.001 r_h \).

3. **Convert the Difference in Radius to Diameter:**
- The difference in diameter \( \Delta d = 2 \Delta r = 2 \times 0.001 r_h = 0.002 r_h \).

4. **Thermal Expansion Formula:**
- The change in diameter due to temperature change can be expressed as:
\[ \Delta d = d_0 \times \alpha \times \Delta T \]
- Where \( d_0 \) is the original diameter, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.

5. **Coefficients of Linear Expansion:**
- Coefficient of linear expansion for aluminum, \( \alpha_a = 24 \times 10^{-6} \, \text{°C}^{-1} \)
- Coefficient of linear expansion for steel, \( \alpha_s = 11 \times 10^{-6} \, \text{°C}^{-1} \)

6. **Calculate the Effective Change in Diameter:**
- The effective change in diameter due to temperature change for both materials:
\[ \Delta d = \Delta T (\alpha_a - \alpha_s) \times d_0 \]
- Rearranging gives:
\[ \Delta T = \frac{\Delta d}{d_0 (\alpha_a - \alpha_s)} \]

7. **Substituting Values:**
- Assume \( d_0 \) is the diameter of the hole.
- Substitute \( \Delta d = 0.002 r_h \), \( \alpha_a = 24 \times 10^{-6} \), and \( \alpha_s = 11 \times 10^{-6} \):
\[ \Delta T = \frac{0.002 r_h}{r_h (24 \times 10^{-6} - 11 \times 10^{-6})} \]
- Simplifying:
\[ \Delta T = \frac{0.002}{13 \times 10^{-6}} \approx 153.85 \, \text{°C} \]

8. **Calculate Heat Required for Steel Sphere:**
- Using \( Q = mc\Delta T \):
- Specific heat of steel \( c_s = 480 \, \text{J/kg°C} \)
- Heat required for steel sphere:
\[ Q_s = m_s \cdot c_s \cdot \Delta T = 1.5 \cdot 480 \cdot 153.85 \approx 111,600 \, \text{J} \]

9. **Calculate Heat Required for Aluminum Plate:**
- Specific heat of aluminum \( c_a = 900 \, \text{J/kg°C} \)
- Heat required for aluminum plate:
\[ Q_a = m_a \cdot c_a \cdot \Delta T = 0.85 \cdot 900 \cdot 153.85 \approx 117,000 \, \text{J} \]

10. **Total Heat Required:**
- Total heat \( Q_{total} = Q_s + Q_a \):
\[ Q_{total} = 111,600 + 117,000 \approx 228,600 \, \text{J} \]

### Final Answer:
The total heat that must be put into the steel sphere and the aluminum plate so that the sphere just passes through the hole is approximately **228,600 Joules**.