Solution
We need to find the length of the cylinder submerged in water, let’s call this length $h_{sub}$.
Step 1: Determine Water Height ($H_w$)
The piston in the bottom tube is held in equilibrium by a horizontal force $F$. This force balances the hydrostatic pressure at the bottom of the vessel.
$$ P_{bottom} \cdot S = F $$
$$ (\rho_w g H_w) S = F \implies H_w = \frac{F}{\rho_w g S} $$
Step 2: Analysis of the Cylinder and Spring
Initial State (Empty Vessel): The base of the cylinder is at height $h_c$. The spring is extended by some amount $x_0$ such that $kx_0 = Mg$.
Final State (Water Added): Buoyant force $F_B$ acts upward on the cylinder. This reduces the extension in the spring, causing the cylinder to move upward by a distance $\Delta y$.
The new force balance is:
$$ k(x_0 – \Delta y) + F_B = Mg $$
Since $kx_0 = Mg$, this simplifies to:
$$ F_B = k \Delta y \quad \text{…(i)} $$
Step 3: Geometry of Submersion
The length of the cylinder submerged, $h_{sub}$, is the difference between the water level and the new position of the cylinder’s base.
- Water Level: $H_w$
- New Base Position: $h_c + \Delta y$ (since it moved up by $\Delta y$)
$$ h_{sub} = H_w – (h_c + \Delta y) \quad \text{…(ii)} $$
Step 4: Solving for $h_{sub}$
From (i), substitute $F_B = \rho_w S_c h_{sub} g$:
$$ \rho_w S_c h_{sub} g = k \Delta y \implies \Delta y = \frac{\rho_w S_c g}{k} h_{sub} $$
Substitute this into (ii):
$$ h_{sub} = H_w – h_c – \frac{\rho_w S_c g}{k} h_{sub} $$
Group $h_{sub}$ terms:
$$ h_{sub} \left( 1 + \frac{\rho_w S_c g}{k} \right) = H_w – h_c $$
$$ h_{sub} \left( \frac{k + \rho_w S_c g}{k} \right) = H_w – h_c $$
$$ h_{sub} = \frac{k(H_w – h_c)}{k + \rho_w S_c g} $$
Finally, substitute $H_w = \frac{F}{\rho_w g S}$:
$$ h_{sub} = \frac{k \left( \frac{F}{\rho_w g S} – h_c \right)}{k + \rho_w S_c g} $$