A metal ball immersed in alcohol weighs `w_(1) " at" 0^(@)C " and" w_(2) " at " 59^(@)C`. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that

Options

A. `w_(1) gt w_(2)`
B. `w_(1)=w_(2)`
C. `w_(1) lt w_(2)`
D. `w_(1)=(w_(1)//2)`

Explanation

To solve the problem, we need to analyze the behavior of a metal ball immersed in alcohol at two different temperatures, 0°C and 59°C. We will use the principles of buoyancy and thermal expansion to derive the relationship between the weights of the ball at these two temperatures.

### Step-by-Step Solution:

1. **Understanding the Apparent Weight**:
- The apparent weight of the ball when immersed in a fluid is given by:
\[ W_{\text{apparent}} = W_{\text{actual}} - F_{\text{upthrust}} \]
- Here, \(W_{\text{actual}}\) is the actual weight of the ball, and \(F_{\text{upthrust}}\) is the upthrust force acting on the ball due to the displaced liquid.

2. **Calculating the Upthrust Force**:
- The upthrust force can be expressed as:
\[ F_{\text{upthrust}} = \text{Volume of the ball} \times \text{Density of the liquid} \times g \]
- For a sphere, the volume is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the upthrust force becomes:
\[ F_{\text{upthrust}} = \frac{4}{3} \pi r^3 \rho_l g \]
- Where \(\rho_l\) is the density of the liquid (alcohol) and \(g\) is the acceleration due to gravity.

3. **Case 1: At 0°C**:
- Let the radius of the ball at 0°C be \(r_0\).
- The upthrust force at 0°C is:
\[ F = \frac{4}{3} \pi r_0^3 \rho_l g \]
- Thus, the apparent weight \(W_1\) at 0°C can be expressed as:
\[ W_1 = W_{\text{actual}} - F = W_{\text{actual}} - \frac{4}{3} \pi r_0^3 \rho_l g \]

4. **Case 2: At 59°C**:
- As the temperature increases, the ball expands. The radius of the ball at 59°C can be expressed as:
\[ r = r_0 (1 + \gamma_s \Delta T) \]
- Where \(\gamma_s\) is the coefficient of cubical expansion of the metal, and \(\Delta T = 59 - 0 = 59\).
- The density of the alcohol at 59°C decreases, which can be expressed as:
\[ \rho_l' = \frac{\rho_l}{1 + \gamma_l \Delta T} \]
- Where \(\gamma_l\) is the coefficient of cubical expansion of the liquid.

5. **Calculating the Upthrust Force at 59°C**:
- The upthrust force at 59°C becomes:
\[ F' = \frac{4}{3} \pi (r_0 (1 + \gamma_s \Delta T))^3 \rho_l' g \]
- Substituting for \(\rho_l'\):
\[ F' = \frac{4}{3} \pi (r_0 (1 + \gamma_s \Delta T))^3 \left(\frac{\rho_l}{1 + \gamma_l \Delta T}\right) g \]

6. **Comparing the Upthrust Forces**:
- Since \(\gamma_s < \gamma_l\), the expansion of the alcohol is greater than that of the metal, leading to a decrease in upthrust force \(F'\) compared to \(F\).
- Therefore, we can conclude that:
\[ F' < F \]

7. **Determining the Apparent Weights**:
- From the expressions for apparent weights:
\[ W_2 = W_{\text{actual}} - F' \]
- Since \(F' < F\), it follows that:
\[ W_2 > W_1 \]

### Conclusion:
Thus, we conclude that the weight of the metal ball at 59°C is greater than at 0°C:
\[ W_2 > W_1 \]