Solution
Concept: When an object is glued to the bottom of a container, the liquid cannot exert upward pressure on the area that is glued. Therefore, the standard buoyancy force (which assumes the object is completely surrounded by fluid) is reduced by the force due to pressure acting on the bottom contact area.
Let $F_{net}$ be the net upward force exerted by the water on the block. Since the base area $A$ is glued and not in contact with water, we subtract the pressure force that would act on this area from the standard buoyancy: $$F_{net\_water} = F_{Buoyancy} – P_{bottom} \times A$$
Here:
- $F_{Buoyancy} = V \rho_w g$
- $P_{bottom} = \rho_w g h$ (Hydrostatic pressure at depth $h$)
For the block to remain attached, the glue must provide a downward force $F_{glue}$ to counteract the net upward pull. The equilibrium equation is: $$F_{net\_water} = mg + F_{glue}$$ $$V \rho_w g – \rho_w g h A = V \rho_o g + F_{glue}$$
We are given the maximum force the glue can withstand is $F_{glue} = 2000 \text{ N}$. Substituting the values:
- $V = 0.5 \text{ m}^3$
- $\rho_w = 1000 \text{ kg/m}^3$
- $\rho_o = 500 \text{ kg/m}^3$
- $A = 100 \text{ cm}^2 = 10^{-2} \text{ m}^2$
- $g = 10 \text{ m/s}^2$
$$ (0.5)(1000)(10) – (1000)(10)(h)(10^{-2}) = (0.5)(500)(10) + 2000 $$ $$ 5000 – 100h = 2500 + 2000 $$ $$ 5000 – 100h = 4500 $$ $$ 100h = 500 $$ $$ h = 5 \text{ m} $$
The minimum (not maximum) height to which water can be filled is 5 m.
Answer: (a)