Solution to Question 6
The system consists of two interconnected vessels. Initially, Vessel A contains a block of ice of height $l_0 = 100$ cm ($1$ m) with an embedded relaxed spring. Vessel B contains water up to height $h_0 = 70$ cm.
Initial Check: The density of ice is $\rho_i = 900$ kg/m$^3$ and water is $\rho_w = 1000$ kg/m$^3$. For an ice block of height $100$ cm to float freely, it requires a draft of $90$ cm. Since the available water head from Vessel B is only $70$ cm, the water pressure is insufficient to lift the ice block. Thus, initially, the ice block rests on the bottom of Vessel A, sealing the connection. There is effectively no water in Vessel A ($h_{A,initial} = 0$).
The ice melts uniformly, reducing its thickness to $0.5$ times the initial value.
New ice thickness: $l_{new} = 0.5 \times 1.0 = 0.5$ m.
New ice mass: $M’_{ice} = \rho_i A l_{new}$.
The melted ice converts to water. By conservation of mass, the total volume of water in the final system is the initial water volume (in B) plus the volume of water from melted ice.
$$ V_{water, final} = V_{water, initial} + V_{melt} $$ $$ A(h_A + h_B) = A h_0 + \frac{\text{Mass of melted ice}}{\rho_w} $$ $$ A(h_A + h_B) = A h_0 + \frac{\rho_i A (0.5 l_0)}{\rho_w} $$
Canceling $A$ and substituting values ($S_{ice} = 0.9$): $$ h_A + h_B = 0.7 + 0.9(0.5) $$ $$ h_A + h_B = 0.7 + 0.45 = 1.15 \text{ m} \quad \dots(1) $$
This is the critical physics step. The spring was initially “embedded and relaxed” within the ice block of length $l_0$. This means the spring coils conform to the ice structure.
When the ice melts from the bottom (implied by the uniform thickness reduction and “extended portion” hint), the bottom half of the spring is liberated into the water, while the top half remains frozen inside the remaining ice block.
- Frozen Part (Top): Remains inside the ice ($l_{new} = 0.5 l_0$). Since the ice holds it rigid at its original relaxed spacing, this part exerts no force.
- Active Part (Bottom): The liberated section corresponds to half the original spring.
- Natural length of active part: $L_{nat} = 0.5 l_0 = 0.5$ m.
- Stiffness of active part: $k’ = 2k$ (since stiffness is inversely proportional to length). $k’ = 2 \times 2500 = 5000$ N/m.
- Current length of active part: Equal to the water depth $h_A$.
The extension of the active spring is $x = L_{current} – L_{nat} = h_A – 0.5$.
The spring force exerted on the piston/ice system is:
$$ F_{spring} = -k’ x = -2k (h_A – 0.5) = 5000(0.5 – h_A) $$
(Positive value indicates an upward push if $h_A < 0.5$).
The fluids in vessels A and B are in equilibrium. We equate the fluid pressures at the bottom connection level.
Vessel B: $P_B = P_{atm} + \rho_w g h_B$.
Vessel A: The pressure at the bottom is the pressure directly under the piston/ice assembly plus the hydrostatic pressure of the water column $h_A$.
The piston and ice float on the water. The total downward force acting on the water surface (interface) is the weight of the remaining ice minus the upward spring force supporting it.
$$ P_{interface} = P_{atm} + \frac{W’_{ice} – F_{spring\_up}}{A} $$
$$ P_A = P_{interface} + \rho_w g h_A $$
Equating $P_A = P_B$ (gauge pressures): $$ \rho_w g h_B = \frac{M’_{ice}g – F_{spring\_up}}{A} + \rho_w g h_A $$ Substituting $F_{spring\_up} = 5000(0.5 – h_A)$: $$ (1000)(10) h_B = \frac{900(0.5 \times 0.5)(10) – 5000(0.5 – h_A)}{0.5} + 10000 h_A $$ $$ 10000 h_B = \frac{2250 – 2500 + 5000 h_A}{0.5} + 10000 h_A $$ $$ 10000 h_B = 2(-250 + 5000 h_A) + 10000 h_A $$ $$ 10000 h_B = -500 + 10000 h_A + 10000 h_A $$ Dividing by 100: $$ 100 h_B = -5 + 200 h_A \quad \dots(2) $$
From (1): $h_A = 1.15 – h_B$. Substitute into (2): $$ 100 h_B = -5 + 200(1.15 – h_B) $$ $$ 100 h_B = -5 + 230 – 200 h_B $$ $$ 300 h_B = 225 $$ $$ h_B = \frac{225}{300} = \frac{3}{4} = 0.75 \text{ m} $$
Final Answer: The height of the water level in vessel B is 75 cm.