A metal ball immersed in alcohol weights `W_1` at `0^@C` and `W_2` at `50^@C`. The coefficient of expansion of cubical the metal is less than that of the alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that

Options

A. W1>W2
B. W2>W1
C. W1=W2
D. None of these

Explanation

To solve the problem, we need to analyze the weights of the metal ball at two different temperatures, 0°C and 50°C, while considering the effects of thermal expansion and buoyancy.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- At 0°C, the weight of the metal ball is \( W_1 \).
- At 50°C, the weight of the metal ball is \( W_2 \).
- The coefficient of volume expansion of the alcohol is greater than that of the metal ball.
...