A metal ball immersed in alchohol weighs W1 at 0 deg C and W2 at 59 deg C. The coefficient at cubical expansion of the metal is less than that of alchol. Assuming that the density of the metal is large compared to that of alchol, it can be shown that : a) W1>W2 b) W1=W2 c) W1

To tackle the problem of the metal ball immersed in alcohol at two different temperatures, we need to consider the principles of buoyancy and thermal expansion. The key factors here are the weights of the ball in alcohol at two temperatures, the coefficients of cubical expansion of both the metal and the alcohol, and the densities involved. Let's break this down step by step.

Understanding the Scenario

When a metal ball is submerged in a liquid, it experiences a buoyant force that is equal to the weight of the liquid displaced by the ball. The weight of the ball in the liquid (W) can be expressed as:

• W = Weight of the ball - Buoyant force

The buoyant force depends on the density of the liquid and the volume of the ball. As the temperature changes, both the density of the alcohol and the volume of the metal ball change due to thermal expansion.

Key Variables

Let’s define some variables:

• W1: Weight of the ball in alcohol at 0°C
• W2: Weight of the ball in alcohol at 59°C
• β_m: Coefficient of cubical expansion of the metal
• β_a: Coefficient of cubical expansion of alcohol
• ρ_m: Density of the metal
• ρ_a: Density of alcohol

Analyzing the Changes

As the temperature increases from 0°C to 59°C, the following occurs:

• The volume of the metal ball increases due to its coefficient of cubical expansion (β_m).
• The density of the alcohol decreases as it warms up, which means it displaces less weight.

Calculating the Weights

At 0°C, the buoyant force (B1) acting on the ball can be expressed as:

• B1 = V_m * ρ_a(0°C)

At 59°C, the buoyant force (B2) becomes:

• B2 = V_m' * ρ_a(59°C)

Where V_m is the initial volume of the metal ball and V_m' is the volume after expansion. The weight of the ball in alcohol at both temperatures can be expressed as:

• W1 = W_m - B1
• W2 = W_m - B2

Comparing W1 and W2

Since the coefficient of cubical expansion of the metal (β_m) is less than that of alcohol (β_a), the volume of the alcohol increases more significantly than that of the metal. This means that:

• The buoyant force B2 at 59°C will be less than B1 at 0°C because the density of alcohol decreases more than the volume increase of the metal ball.

Thus, we can conclude that:

• W2 = W_m - B2 < W_m - B1 = W1

Final Conclusion

From the analysis, we find that W1 is greater than W2, leading us to the conclusion:

• W1 > W2

Therefore, the correct answer to the question is option (a) W1 > W2. This demonstrates how temperature changes can affect buoyancy and weight in fluids, particularly when dealing with materials that expand differently.