© All Rights reserved @ LearnWithDash
Step-by-step Solution:
Step 1: Identify the governing principle
The problem deals with a cylinder floating in water. According to the principle of flotation, the weight of the floating body is balanced by the buoyant force exerted by the fluid.
Step 2: Express the buoyant force at 0°C
Let $A$ be the cross-sectional area of the cylinder (constant) and $g$ be the acceleration due to gravity. At 0°C, the height of the submerged part of the cylinder is $80\text{ cm}$ (since the total length is $100\text{ cm}$ and $20\text{ cm}$ is above the water surface). The density of water at $0^\circ \text{C}$ is $\rho_{0^\circ C}$. The buoyant force $F_{B1}$ equals the weight of the displaced water:
$F_{B1} = \rho_{0^\circ C} \times \bigl(A \times 80\bigr) \times g.$
Let $m$ be the mass of the cylinder. Hence,
$\rho_{0^\circ C} \times \bigl(A \times 80\bigr) \times g = m\,g.$
From this, we get
$m = \rho_{0^\circ C}\,A \times 80.$
Step 3: Express the buoyant force at 4°C
When the temperature is increased to $4^\circ \text{C}$, the height of the cylinder above the water surface is $21\text{ cm}$, so the submerged part is $79\text{ cm}$. Let the density of water at $4^\circ\text{C}$ be $\rho_{4^\circ C}$. The buoyant force $F_{B2}$ in this scenario is:
$F_{B2} = \rho_{4^\circ C} \times \bigl(A \times 79\bigr) \times g.$
Because the cylinder still floats, its weight is the same and must equal this buoyant force:
$\rho_{4^\circ C} \times \bigl(A \times 79\bigr) \times g = m\,g.$
Thus,
$m = \rho_{4^\circ C}\,A \times 79.$
Step 4: Relate both scenarios to find the density ratio
From step 2 and step 3, we have expressions for $m$. Equating them, we get:
$\rho_{4^\circ C}\,A \times 79 = \rho_{0^\circ C}\,A \times 80.$
Canceling $A$ on both sides and rearranging:
$\frac{\rho_{4^\circ C}}{\rho_{0^\circ C}} = \frac{80}{79}.$
Numerically, $\frac{80}{79} \approx 1.01.$
Step 5: Conclusion
Therefore, the density of water at $4^\circ \text{C}$ relative to the density at $0^\circ \text{C}$ is about $1.01.$
