做上面的第二小题---**Problem Description:**
7. As shown in Figure 甲, a uniform, non-absorbent cylindrical body A weighs 10N, has a height of 20cm, and a base area of 100cm². As shown in Figure 乙, a thin-walled cylindrical container B of negligible mass is placed on a horizontal table, with a base area of 200cm² and a height of 32cm. Now, the center of the bottom surface of cylinder A is connected to the bottom of container B with a 6cm long thin string (ignore the mass, volume, etc. of the thin string), as shown in Figure 丙. A is placed in the center of container B and is stationary. Water is slowly injected into container B at a constant rate of 100cm³ per minute. After injecting water for 12 minutes for the first time, the injection is paused. (Ignore secondary factors such as the object absorbing liquid. It is known that ρ水 = 1.0 × 10³kg/m³, and g is taken as 10N/kg)
**(1) Calculate the pressure exerted by the container on the table when no water is added in Figure 丙.**
**(2) After the first water injection, slowly lift A vertically upwards by 2cm. Calculate the buoyant force on A at this moment.**
**(3) After releasing A and it becomes stationary again, start the second water injection. From the start of the second injection until A is just fully submerged, find the functional relationship between the water pressure p on the container bottom and the injection time t (min).**
**Diagram Description:**
* **Figure 甲:** A vertical cylinder labeled 'A'. It is annotated with "GA=10N", "hA=20cm", and "SA=100cm²".
* **Figure 乙:** A vertical open-top cylinder (container) labeled 'B'. It is annotated with "hB=32cm" and "SB=200cm²".
* **Figure 丙:** Container B placed on a surface. Cylinder A is placed inside B, centered horizontally. A thin line connects the bottom center of A to the bottom of B. The length of this line is labeled "l=6cm". The figures show A is partially submerged in water (this is the state *after* the first injection, based on the problem context, but the label '丙' is below the initial setup description. The question (1) refers to '丙' *before* adding water, which is consistent with A being in B and connected by the string).
**Given Data and Constants:**
* Weight of A (GA): 10N
* Height of A (hA): 20cm
* Base Area of A (SA): 100cm²
* Base Area of B (SB): 200cm²
* Height of B (hB): 32cm
* Length of string (l): 6cm
* Water injection rate: 100cm³/min
* First injection time: 12min
* Density of water (ρ水): 1.0 × 10³kg/m³
* Acceleration due to gravity (g): 10N/kg
**Handwritten Notes/Calculations Present in Image:**
* Calculation related to Question (1):
`p = F/S_B = 10N / (2*10^-2 m^2) = 5 * 10^2 Pa` (Result `5*10^2 Pa` is circled)
Note: Conversion 200cm² = 200 * (10⁻²m)² = 200 * 10⁻⁴m² = 2 * 10⁻²m² is used.
* Calculation related to Question (2):
`V水 = 100cm³/min * 12min = 1200cm³`