An aluminium sphere is dipped into water. Which of the following is true?

The correct option is: Buoyancy will be less in water at 0°C than that in water at 4°C

Step-by-Step Explanation:

1. Understanding Buoyant Force:
According to Archimedes' principle, when an object is immersed in a fluid, it experiences an upward buoyant force ($F_b$) equal to the weight of the fluid displaced by the object. The formula for buoyant force is given by:
$F_b=V\cdot \rho \cdot g$
where:
- $V$ is the volume of the submerged portion of the sphere,
- $\rho$ is the density of the fluid (water in this case), and
- $g$ is the acceleration due to gravity.

2. Anomalous Expansion of Water:
Water exhibits a unique property known as anomalous expansion. Unlike most liquids, which contract continuously as temperature decreases, water reaches its maximum density at $4°\text{C}$.
Specifically:
- The density of water at $4°\text{C}$ (${\rho }_{4°\text{C}}$) is at its maximum value of approximately $1000{\text{kg/m}}^3$.
- As the temperature decreases from $4°\text{C}$ to $0°\text{C}$, water expands, which means its density decreases. Therefore, ${\rho }_{0°\text{C}}<{\rho }_{4°\text{C}}$.

3. Comparing Buoyancy at both temperatures:
Since the aluminum sphere is dipped in water, the volume $V$ of water displaced is essentially determined by the volume of the sphere. Assuming the negligible thermal expansion of the aluminum sphere between $0°\text{C}$ and $4°\text{C}$, the volume $V$ remains virtually constant.
Since the density of water at $0°\text{C}$ is less than the density of water at $4°\text{C}$ (${\rho }_{0°\text{C}}<{\rho }_{4°\text{C}}$), the buoyant force at $0°\text{C}$ will be less than the buoyant force at $4°\text{C}$:
$F_{b(0°\text{C})}<F_{b(4°\text{C})}$
Thus, the buoyancy will be less in water at $0°\text{C}$ than that in water at $4°\text{C}$.