Solution
Analysis: We need to compare the volume of water displaced by the ice while it is solid ($V_{disp}$) versus the volume the ice water occupies after it melts ($V_{melt}$).
1. Initial State (Solid Ice):
The ice is held partially submerged by a string with tension $T = 1.0 \text{ N}$. The forces acting on the ice are Buoyancy ($B$) upwards, Gravity ($mg$) downwards, and Tension ($T$) downwards (since the ice wants to float up).
$$B = mg + T$$
$$V_{disp} \rho_w g = mg + T \implies V_{disp} = \frac{mg + T}{\rho_w g}$$
2. Final State (Melted):
When the ice melts, it turns into water. The mass $m$ remains conserved. The volume of this new water is:
$$V_{melt} = \frac{m}{\rho_w}$$
3. Change in Volume:
The change in the volume occupied in the tank is $\Delta V = V_{melt} – V_{disp}$.
$$ \Delta V = \frac{m}{\rho_w} – \left( \frac{mg + T}{\rho_w g} \right) $$
$$ \Delta V = \frac{m}{\rho_w} – \frac{m}{\rho_w} – \frac{T}{\rho_w g} $$
$$ \Delta V = – \frac{T}{\rho_w g} $$
The negative sign indicates a decrease in total volume, meaning the water level will fall.
Calculation: Given $T = 1.0 \text{ N}$, $\rho_w = 1000 \text{ kg/m}^3$, $A = 400 \text{ cm}^2 = 400 \times 10^{-4} \text{ m}^2$. $$ \text{Level Shift } (\Delta h) = \frac{\Delta V}{A} = \frac{-T}{\rho_w g A} $$ $$ \Delta h = \frac{-1.0}{1000 \times 10 \times (400 \times 10^{-4})} $$ $$ \Delta h = \frac{-1.0}{1000 \times 10 \times 0.04} = \frac{-1}{400} \text{ m} $$ $$ \Delta h = -0.0025 \text{ m} = -2.5 \text{ mm} $$
The water level shifts 2.5 × 10⁻³ m downwards.
Answer: (b)