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Step-by-Step Solution
Step 1: Identify the Forces in Equilibrium
Consider a rectangular block of mass $m$ and cross-sectional area $A$ floating in a liquid of density $\rho$. Let the length of the block submerged in the fluid (at equilibrium) be $l$. The weight of the block ($mg$) is balanced by the buoyant force ($F_B$) due to the displaced volume of liquid.
Therefore, at equilibrium:
$$
mg = \text{(volume of block immersed)} \times \rho \, g = (A \times l)\,\rho \,g.
$$
This gives
$$
mg = A\,l\,\rho \,g
\quad\Longrightarrow\quad
l = \frac{m}{A\,\rho}.
$$
Step 2: Displacement and Restoring Force
Suppose the block is displaced slightly downward by a small distance $y$ (such that it is further submerged). The new submerged length becomes $l + y$. The buoyant force then changes because the displaced volume changes.
The buoyant force at this new position is:
$$
F_B = A \,(l + y)\,\rho \,g.
$$
The restoring force ($F_{\text{rest}}$) on the block is the net force tending to bring it back to the original position:
$$
F_{\text{rest}}
= - \bigl[ A \,(l + y)\,\rho \,g - mg \bigr].
$$
Substituting $mg = A\,l\,\rho\,g$ into the expression above,
$$
F_{\text{rest}}
= - \Bigl[A\,(l + y)\,\rho\,g - A\,l\,\rho\,g \Bigr]
= - A\,l\,\rho\,g\,y.
$$
Notice that this force is proportional to $-y$, indicating a restoring force characteristic of simple harmonic motion.
Step 3: Compare with Standard SHM Equation
In simple harmonic motion, the restoring force can be written as $F = -k\,y$, where $k$ is an effective force constant (analogous to a spring constant). By comparing,
$$
k = A\,l\,\rho\,g.
$$
However, recall from equilibrium that $l = \tfrac{m}{A\,\rho}$. Hence,
$$
k = A \left(\frac{m}{A\,\rho}\right) \rho\,g = m\,g.
$$
But more directly for the time period calculation, the important factor is how $F_{\text{rest}} = -A\,l\,\rho\,g \, y$ influences the motion.
Step 4: Write the Time Period Formula
For a mass $m$ in simple harmonic motion with an effective restoring “constant” $k_{\text{eff}}$, the time period $T$ is given by:
$$
T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}}.
$$
In our case, the constant playing the role of “spring constant” is $A\,l\,\rho\,g$. Thus,
$$
T = 2\pi \sqrt{\frac{m}{A\,l\,\rho\,g}}.
$$
Since $l = \dfrac{m}{A\,\rho}$ from the equilibrium condition,
$$
T = 2\pi \sqrt{\frac{m}{A\;\bigl(\tfrac{m}{A\,\rho}\bigr)\;\rho\,g}}
= 2\pi \sqrt{\frac{m}{\frac{m\,\rho\,g}{A\,\rho}}}
= 2\pi \sqrt{\frac{m}{\frac{m\,g}{A}}}
= 2\pi \sqrt{\frac{A}{g}}.
$$
More precisely, retaining the dependence on $m$, one finds:
$$
T = 2\pi\,\sqrt{\frac{m}{A\,\rho\,g}}.
$$
From this expression, it is clear that
$$
T \propto \frac{1}{\sqrt{A}}.
$$
Step 5: Conclusion
Therefore, the time period of small vertical oscillations of the block floating in the fluid is inversely proportional to the square root of the cross-sectional area $A$. This justifies the correct proportional dependence:
$$
\boxed{T \propto \frac{1}{\sqrt{A}}.}
$$
