Solution Analysis
Consider a rod of length $l$, cross-sectional area $A$, and density $\rho$ hinged at the bottom of a container. Let the density of water be $\rho_w$ (where $\rho_w > \rho$). As the water level $h$ rises, the rod experiences a buoyant force and attempts to float.
Let $l_i$ be the length of the rod immersed in water at any instant, and $\theta$ be the angle the rod makes with the horizontal.
Step 1: Equilibrium in Floating Condition
When the water level rises initially, the rod lifts off the bottom due to buoyancy. The rod is in rotational equilibrium about the hinge. The forces creating torque about the hinge are:
- Gravity ($mg$): Acts downwards at the center of the rod (distance $l/2$ from the hinge).
- Buoyant Force ($F_B$): Acts upwards at the center of the immersed portion (distance $l_i/2$ from the hinge).
Balancing the torques ($\tau_{net} = 0$):
$$ \tau_{gravity} = \tau_{buoyancy} $$ $$ (mg) \cdot \frac{l}{2} \cos\theta = (F_B) \cdot \frac{l_i}{2} \cos\theta $$Substituting $mg = \rho A l g$ and $F_B = \rho_w A l_i g$:
$$ (\rho A l g) \frac{l}{2} = (\rho_w A l_i g) \frac{l_i}{2} $$ $$ \rho l^2 = \rho_w l_i^2 $$Observation: In this floating phase, the immersed length $l_i$ depends only on the densities and the total length. It is constant and independent of the water height $h$ or angle $\theta$. This explains the initial horizontal segment of the graph.
Step 2: Transition to Vertical Position
The water height is related to the immersed length by geometry: $h = l_i \sin\theta$. As $h$ increases, $\theta$ increases. The maximum height for this “oblique floating” phase occurs when the rod becomes vertical ($\theta = 90^\circ, \sin\theta = 1$).
$$ h = l_i = l \sqrt{\frac{\rho}{\rho_w}} $$Up to this value of $h$, $l_i$ remains constant.
Step 3: Vertical and Fully Submerged Phases
Vertical Phase: Once $h > l \sqrt{\frac{\rho}{\rho_w}}$, the rod stays vertical because the buoyant force is sufficient to keep it upright. In this vertical position, the immersed length is simply equal to the depth of the water.
$$ l_i = h $$This represents the linear portion of the graph with a slope of 1 (following the line $y=x$).
Saturation: This linear increase continues until the water level reaches the top of the rod ($h = l$). For any $h > l$, the rod is fully submerged, and the immersed length remains constant at:
$$ l_i = l $$